Quadratic Functions and Equations

Parabolic Anatomy is the study of the important parts of a parabola, which is the graph of a quadratic function.

A quadratic function usually looks like this:

$$y=ax^2+bx+c$$

Its graph is a parabola, a U-shaped curve that can open upward or downward.

When we study parabolic anatomy, we focus on three main features:

• Vertex – the highest or lowest point of the parabola
• Axis of symmetry – the vertical line that splits the parabola into two matching halves
• Concavity – whether the parabola opens up or down

Knowing these features helps us understand the graph quickly, even before we draw it.

1. The Vertex

The vertex is the turning point of the parabola.

• If the parabola opens upward, the vertex is the lowest point.
• If the parabola opens downward, the vertex is the highest point.

If a quadratic is written in vertex form, it is easiest to find the vertex:

$$y=a(x-h)^2+k$$

In this form, the vertex is:

$$ (h,k) $$

So if we have:

$$y=2(x-3)^2+5$$

the vertex is \,((3,5))\,.

2. The Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex and divides the parabola into two equal sides.

If the vertex is \,((h,k))\,, then the axis of symmetry is:

$$x=h$$

For the function

$$y=2(x-3)^2+5$$

the axis of symmetry is:

$$x=3$$

This means the left and right sides of the parabola mirror each other across the line \,x=3\,.

3. Concavity

Concavity tells us which way the parabola opens.

• If \,a>0\,, the parabola opens upward.
• If \,a<0\,, the parabola opens downward.

In the equation

$$y=a(x-h)^2+k$$

the value of \,a\, controls the direction.

Examples:

• In \,y=3(x+1)^2-4\,, \,a=3\,, so it opens up.
• In \,y=-2(x-1)^2+6\,, \,a=-2\,, so it opens down.

4. What the Leading Coefficient Tells Us

The leading coefficient is the number in front of \,x^2\, when the equation is in standard form \,y=ax^2+bx+c\,. It is still the same \,a\, in vertex form.

This number tells us:

• the direction the parabola opens
• how wide or narrow the parabola appears

If \,|a|\, is larger, the parabola is usually narrower.

If \,|a|\, is smaller, the parabola is usually wider.

For example:

• \,y=x^2\, is a basic parabola
• \,y=4x^2\, is narrower
• \,y=\frac{1}{2}x^2\, is wider

5. Finding the Vertex from Standard Form

Sometimes a quadratic is given in standard form:

$$y=ax^2+bx+c$$

To find the x-coordinate of the vertex, use:

$$x=\frac{-b}{2a}$$

Then substitute that x-value back into the equation to find y.

This gives the vertex.

Once you know the vertex, the axis of symmetry is the same x-value.

Worked Example 1: Identify parts from vertex form

Find the vertex, axis of symmetry, and concavity of:

$$y=(x-4)^2+1$$

Step 1: Find the vertex.

This is already in vertex form \,y=a(x-h)^2+k\,.

So \,h=4\, and \,k=1\,.

The vertex is:

$$ (4,1) $$

Step 2: Find the axis of symmetry.

The axis of symmetry is:

$$x=4$$

Step 3: Find the concavity.

Here \,a=1\,, which is positive.

So the parabola opens upward.

Answer:

• Vertex: \,((4,1))\,
• Axis of symmetry: \,x=4\,
• Concavity: opens upward

Worked Example 2: Identify parts from vertex form with a negative coefficient

Find the vertex, axis of symmetry, and concavity of:

$$y=-3(x+2)^2+5$$

Step 1: Find the vertex.

Rewrite \,x+2\, as \,x-(-2)\,.

So \,h=-2\, and \,k=5\,.

The vertex is:

$$(-2,5)$$

Step 2: Find the axis of symmetry.

$$x=-2$$

Step 3: Find the concavity.

Since \,a=-3\,, the parabola opens downward.

Answer:

• Vertex: \,((-2,5))\,
• Axis of symmetry: \,x=-2\,
• Concavity: opens downward

Worked Example 3: Find the vertex from standard form

Find the vertex, axis of symmetry, and concavity of:

$$y=x^2-6x+5$$

Step 1: Identify \,a\, and \,b\,.

Here:

$$a=1 \quad b=-6$$

Step 2: Find the x-coordinate of the vertex.

Use:

$$x=\frac{-b}{2a}$$

Substitute the values:

$$x=\frac{-(-6)}{2(1)}=\frac{6}{2}=3$$

Step 3: Find the y-coordinate.

Substitute \,x=3\, into the equation:

$$y=3^2-6(3)+5$$

$$y=9-18+5=-4$$

So the vertex is:

$$ (3,-4) $$

Step 4: Find the axis of symmetry.

$$x=3$$

Step 5: Find the concavity.

Since \,a=1\,, the parabola opens upward.

Answer:

• Vertex: \,((3,-4))\,
• Axis of symmetry: \,x=3\,
• Concavity: opens upward

Worked Example 4: Standard form with downward opening

Find the vertex, axis of symmetry, and concavity of:

$$y=-2x^2+8x-1$$

Step 1: Identify \,a\, and \,b\,.

$$a=-2 \quad b=8$$

Step 2: Find the x-coordinate of the vertex.

$$x=\frac{-b}{2a}=\frac{-8}{2(-2)}=\frac{-8}{-4}=2$$

Step 3: Find the y-coordinate.

Substitute \,x=2\,:

$$y=-2(2^2)+8(2)-1$$

$$y=-2(4)+16-1$$

$$y=-8+16-1=7$$

So the vertex is:

$$ (2,7) $$

Step 4: Axis of symmetry.

$$x=2$$

Step 5: Concavity.

Since \,a=-2\,, the parabola opens downward.

Answer:

• Vertex: \,((2,7))\,
• Axis of symmetry: \,x=2\,
• Concavity: opens downward

Helpful Tips

• The vertex is always on the axis of symmetry.
• The sign of \,a\, tells the direction: positive is up, negative is down.
• In vertex form, the vertex is easy to read directly.
• In standard form, use \,x=\frac{-b}{2a}\, to start.

Common Mistakes to Avoid

• Mixing up the sign in vertex form. In \,y=(x-3)^2+2\,, the vertex is \,((3,2))\,, not \,((-3,2))\,.
• Forgetting that \,x+4\, means \,x-(-4)\,, so the vertex x-value is \,-4\,.
• Using the wrong sign for concavity. Always check whether \,a\, is positive or negative.
• Finding \,x=\frac{-b}{2a}\, but forgetting to plug it back in to get the y-value.

Summary

A parabola is the graph of a quadratic function. Its main parts are the vertex, the axis of symmetry, and its concavity.

In vertex form \,y=a(x-h)^2+k\,, the vertex is \,((h,k))\, and the axis of symmetry is \,x=h\,. The value of \,a\, tells whether the parabola opens up or down, and it also affects how wide or narrow the graph looks.

In standard form \,y=ax^2+bx+c\,, you can find the vertex using:

$$x=\frac{-b}{2a}$$

Then substitute that x-value into the equation to find y. Once you can identify these parts, you can understand the shape and position of any parabola much more easily.

Vertex Form Analysis is about reading important information from a quadratic function when it is written in the form

$$f(x)=a(x-h)^2+k$$

This form is called vertex form. It is very useful because it tells us the vertex, whether the parabola opens up or down, and the function’s maximum or minimum value.

Quadratic functions graph as parabolas. A parabola is a curved graph shaped like a “U” or an upside-down “U”. The vertex is the turning point of the parabola.

In this lesson, you will learn how to look at a function in vertex form and quickly answer questions about its graph.

1. The parts of vertex form

The standard vertex form is

$$f(x)=a(x-h)^2+k$$

Each part has a meaning:

• a ext{ controls the direction and width}
• h ext{ gives the x-coordinate of the vertex}
• k ext{ gives the y-coordinate of the vertex}

So the vertex is always

$$\boxed{(h,k)}$$

This is one of the most important ideas in this topic.

Be careful: the value of h comes from the expression \((x-h)\). That means the sign inside the bracket is the opposite of the x-coordinate of the vertex.

For example:

• In \((x-4)^2\), the vertex x-coordinate is \(4\)
• In \((x+3)^2\), the vertex x-coordinate is \(-3\) because \(x+3=x-(-3)\)

2. How the value of \(a\) changes the graph

The number \(a\) tells us whether the parabola opens upward or downward.

• If \(a>0\), the parabola opens upward
• If \(a<0\), the parabola opens downward

This helps us know if the vertex is a minimum or a maximum.

• If the parabola opens upward, the vertex is the lowest point, so it is a minimum
• If the parabola opens downward, the vertex is the highest point, so it is a maximum

The value of \(a\) also affects how narrow or wide the parabola looks.

• If \(|a|>1\), the parabola is narrower
• If \(0<|a|<1\), the parabola is wider

For vertex form analysis in 9th Grade, the most important job of \(a\) is usually to tell the direction and whether there is a maximum or minimum.

3. Axis of symmetry

Every parabola has a vertical line that cuts it into two matching halves. This line is called the axis of symmetry.

For a quadratic in vertex form, the axis of symmetry is

$$\boxed{x=h}$$

This makes sense because the vertex sits on this line.

For example, if the vertex is \((2,-1)\), then the axis of symmetry is \(x=2\).

4. Maximum and minimum values

The y-value of the vertex tells the extreme value of the function.

• If the parabola opens upward, the minimum value is \(k\)
• If the parabola opens downward, the maximum value is \(k\)

This happens at \(x=h\).

So for \(f(x)=a(x-h)^2+k\):

• Vertex: \((h,k)\)
• Axis of symmetry: \(x=h\)
• Extreme value: \(k\)

5. A step-by-step method for analyzing vertex form

When you are given a quadratic in vertex form, use these steps:

1. Identify \(a\), \(h\), and \(k\)
2. Write the vertex \((h,k)\)
3. Write the axis of symmetry \(x=h\)
4. Check the sign of \(a\) to see if it opens up or down
5. State whether the function has a maximum or minimum
6. Give the extreme value, which is \(k\)

Worked Example 1

Analyze the function

$$f(x)=(x-3)^2+2$$

Step 1: Identify the parts.

Here, \(a=1\), \(h=3\), and \(k=2\).

Step 2: Find the vertex.

The vertex is

$$ (h,k)=(3,2) $$

Step 3: Find the axis of symmetry.

$$x=3$$

Step 4: Decide whether it opens up or down.

Since \(a=1>0\), the parabola opens upward.

Step 5: State the extreme value.

Because it opens upward, the vertex is a minimum.

The minimum value is \(2\), and it happens when \(x=3\).

Answer:

• Vertex: \((3,2)\)
• Axis of symmetry: \(x=3\)
• Opens upward
• Minimum value: \(2\)

Worked Example 2

Analyze the function

$$g(x)=-2(x+1)^2+5$$

Step 1: Identify the parts.

We can rewrite \((x+1)^2\) as \((x-(-1))^2\), so:

$$a=-2, \quad h=-1, \quad k=5$$

Step 2: Find the vertex.

$$ (h,k)=(-1,5) $$

Step 3: Find the axis of symmetry.

$$x=-1$$

Step 4: Decide whether it opens up or down.

Since \(a=-2<0\), the parabola opens downward.

Step 5: State the extreme value.

Because it opens downward, the vertex is a maximum.

The maximum value is \(5\), and it happens when \(x=-1\).

Answer:

• Vertex: \((-1,5)\)
• Axis of symmetry: \(x=-1\)
• Opens downward
• Maximum value: \(5\)

Worked Example 3

Analyze the function

$$y=\frac{1}{2}(x-4)^2-3$$

Step 1: Identify the parts.

$$a=\frac{1}{2}, \quad h=4, \quad k=-3$$

Step 2: Find the vertex.

$$ (4,-3) $$

Step 3: Find the axis of symmetry.

$$x=4$$

Step 4: Decide whether it opens up or down.

Since \(a=\frac{1}{2}>0\), it opens upward.

Step 5: State the extreme value.

The vertex is a minimum, so the minimum value is \(-3\).

This happens when \(x=4\).

Extra observation: because \(|a|=\frac{1}{2}<1\), this parabola is wider than \(y=x^2\).

Answer:

• Vertex: \((4,-3)\)
• Axis of symmetry: \(x=4\)
• Opens upward
• Minimum value: \(-3\)

Worked Example 4

Analyze the function

$$f(x)=-3(x-2)^2-7$$

Step 1: Identify the parts.

$$a=-3, \quad h=2, \quad k=-7$$

Step 2: Find the vertex.

$$ (2,-7) $$

Step 3: Find the axis of symmetry.

$$x=2$$

Step 4: Decide whether it opens up or down.

Since \(a=-3<0\), it opens downward.

Step 5: State the extreme value.

The vertex is a maximum, so the maximum value is \(-7\).

This happens when \(x=2\).

Extra observation: because \(|a|=3>1\), the parabola is narrower than \(y=x^2\).

Answer:

• Vertex: \((2,-7)\)
• Axis of symmetry: \(x=2\)
• Opens downward
• Maximum value: \(-7\)

6. Common mistakes to avoid

• Forgetting the sign change in \(h\)
  In \((x+4)^2\), the vertex x-coordinate is \(-4\), not \(4\).
• Mixing up maximum and minimum
  If \(a>0\), the parabola opens up and has a minimum. If \(a<0\), it opens down and has a maximum.
• Using \(a\) as part of the vertex
  The vertex is \((h,k)\), not \((a,k)\).
• Missing the axis of symmetry
  The axis of symmetry is always \(x=h\).

7. Quick practice ideas

Try analyzing these on your own:

• \(y=(x+2)^2-6\)
• \(y=-\frac{1}{3}(x-5)^2+1\)
• \(y=4(x-1)^2\)

For each one, find:

• the vertex
• the axis of symmetry
• whether it opens up or down
• the maximum or minimum value

8. Summary

Vertex form makes quadratics easier to understand because the most important graph features are already built into the equation.

In \(f(x)=a(x-h)^2+k\):

• The vertex is \((h,k)\)
• The axis of symmetry is \(x=h\)
• If \(a>0\), the parabola opens up and has a minimum
• If \(a<0\), the parabola opens down and has a maximum
• The extreme value is the y-value of the vertex, which is \(k\)

If you can identify \(a\), \(h\), and \(k\), then you can quickly describe the graph of a quadratic in vertex form.

Solving Quadratics by Factoring

A quadratic equation is an equation where the highest power of the variable is 2. A common form is:

$$ax^2+bx+c=0$$

In this lesson, you will learn how to solve quadratic equations by factoring. This method helps you find the values of the variable that make the equation true.

These solutions are also called the roots, zeros, or x-intercepts of the quadratic when it is graphed.

Why factoring works

Factoring rewrites a quadratic as a product of two simpler expressions. For example,

$$x^2+5x+6=(x+2)(x+3)$$

If a product equals zero, then at least one factor must equal zero. This is called the Zero Product Property.

$$ab=0 \Rightarrow a=0 \text{ or } b=0$$

This property is the key to solving quadratics by factoring.

Steps for solving by factoring

1. Make sure the equation is set equal to 0.

2. Factor the quadratic expression completely.

3. Set each factor equal to 0.

4. Solve each simple equation.

5. Check your solutions if needed.

Important idea: You can only use the Zero Product Property when one side of the equation is 0.

For example, if you have

$$x^2+7x+10=3$$

you should first move everything to one side:

$$x^2+7x+7=0$$

Then try to factor.

How to factor simple quadratics

When the leading coefficient is 1, like in

$$x^2+bx+c$$

look for two numbers that:

• multiply to give \(c\)

• add to give \(b\)

For example, to factor \(x^2+7x+12\), we need two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4.

So,

$$x^2+7x+12=(x+3)(x+4)$$

Worked Example 1

Solve:

$$x^2+5x+6=0$$

Step 1: Factor

We need two numbers that multiply to 6 and add to 5. They are 2 and 3.

$$x^2+5x+6=(x+2)(x+3)$$

Step 2: Use the Zero Product Property

$$ (x+2)(x+3)=0 $$

So,

$$x+2=0 \quad \text{or} \quad x+3=0$$

Step 3: Solve each equation

$$x=-2 \quad \text{or} \quad x=-3$$

Answer: The solutions are \(-2\) and \(-3\).

If you graphed the parabola \(y=x^2+5x+6\), it would cross the x-axis at \((-2,0)\) and \((-3,0)\).

Worked Example 2

Solve:

$$x^2-9=0$$

This is a special type of quadratic called a difference of squares.

$$x^2-9=x^2-3^2=(x-3)(x+3)$$

Now use the Zero Product Property:

$$ (x-3)(x+3)=0 $$

Set each factor equal to 0:

$$x-3=0 \quad \text{or} \quad x+3=0$$

So,

$$x=3 \quad \text{or} \quad x=-3$$

Answer: The solutions are \(3\) and \(-3\).

Worked Example 3

Solve:

$$x^2+x-12=0$$

We need two numbers that multiply to \(-12\) and add to \(1\).

Those numbers are 4 and \(-3\), because

$$4\cdot(-3)=-12 \quad \text{and} \quad 4+(-3)=1$$

So,

$$x^2+x-12=(x+4)(x-3)$$

Now set each factor equal to 0:

$$x+4=0 \quad \text{or} \quad x-3=0$$

Solving gives:

$$x=-4 \quad \text{or} \quad x=3$$

Answer: The solutions are \(-4\) and \(3\).

Worked Example 4

Solve:

$$2x^2+7x+3=0$$

This quadratic has a leading coefficient that is not 1, so factoring takes a little more care.

We want to factor it into two binomials:

$$2x^2+7x+3=(2x+1)(x+3)$$

You can check by multiplying:

$$ (2x+1)(x+3)=2x^2+6x+x+3=2x^2+7x+3 $$

Now use the Zero Product Property:

$$ (2x+1)(x+3)=0 $$

Set each factor equal to 0:

$$2x+1=0 \quad \text{or} \quad x+3=0$$

Solve each one:

$$2x=-1 \Rightarrow x=-\frac{1}{2}$$

$$x=-3$$

Answer: The solutions are \(-\frac{1}{2}\) and \(-3\).

When the equation is not equal to 0 first

Sometimes you must rearrange before factoring.

For example, solve:

$$x^2+4x=12$$

Move 12 to the left side:

$$x^2+4x-12=0$$

Now factor. We need two numbers that multiply to \(-12\) and add to 4. They are 6 and \(-2\).

$$x^2+4x-12=(x+6)(x-2)$$

Set each factor equal to 0:

$$x+6=0 \quad \text{or} \quad x-2=0$$

So,

$$x=-6 \quad \text{or} \quad x=2$$

Common mistakes to avoid

• Not setting the equation equal to 0 first. You must do this before using the Zero Product Property.

• Factoring incorrectly. Always check that your factors multiply back to the original quadratic.

• Forgetting to set both factors equal to 0. Each factor can lead to a solution.

• Sign errors. Be careful with positive and negative numbers when finding factor pairs.

How to check your answers

Substitute each solution back into the original equation.

For Example 1, one solution was \(x=-2\):

$$(-2)^2+5(-2)+6=4-10+6=0$$

Since it makes the equation true, \(x=-2\) is correct.

Quick review of the full process

1. Write the equation so one side is 0.

2. Factor the quadratic completely.

3. Use the Zero Product Property.

4. Solve for the variable.

5. Check your answers.

Summary

Solving quadratics by factoring means rewriting a quadratic equation as a product of factors and then using the Zero Product Property.

When a quadratic is written in the form \(ax^2+bx+c=0\), you factor it, set each factor equal to 0, and solve. This gives the exact solutions, which are also the x-intercepts of the graph.

With practice, you will get faster at spotting the numbers needed to factor and solve correctly.

Completing the Square is a method for rewriting a quadratic expression so that part of it becomes a perfect square. This helps us solve quadratic equations and also rewrite a quadratic function in a form that makes its graph easier to understand.

A quadratic expression usually looks like \(x^2+bx+c\), and a quadratic equation might look like \(x^2+bx+c=0\). When we complete the square, we turn the first two terms into something like \((x+a)^2\).

This works because perfect square binomials follow a pattern:

For example:

• \((x+3)^2=x^2+6x+9\)
• \((x-5)^2=x^2-10x+25\)

Notice something important: the middle term is always twice the number inside the brackets, and the last term is that number squared.

This idea is the key to completing the square.

Why do we complete the square?

• To solve quadratic equations
• To rewrite a quadratic in a useful form
• To find the turning point of a parabola
• To understand the shape and position of a graph

Let us start with the basic case where the coefficient of \(x^2\) is 1.

Suppose we have:

We want to turn this into the start of a perfect square. We ask: what number goes with \(x\) in \((x+?)^2\)?

Since \((x+a)^2=x^2+2ax+a^2\), the middle term \(8x\) means:

So:

That means:

We had to add 16 to make a perfect square, so we also subtract 16 to keep the expression equal to the original one.

The rule for expressions like \(x^2+bx\) is:

1. Take half of the coefficient of \(x\)
2. Square it
3. Add and subtract that number

In symbols:

This may look complicated, but the idea is simple: half the middle number, then square it.

Worked Example 1: Rewrite as a completed square

Rewrite \(x^2+10x+7\) by completing the square.

Step 1: Look at the \(x\)-term. The coefficient is 10.

Step 2: Half of 10 is 5.

Step 3: Square 5 to get 25.

Add and subtract 25:

Group the perfect square:

So the completed square form is:

Worked Example 2: Solve a quadratic equation by completing the square

Solve:

Step 1: Move the constant to the other side.

Step 2: Complete the square.

Half of 6 is 3, and \(3^2=9\). Add 9 to both sides:

Step 3: Take the square root of both sides.

Step 4: Solve for \(x\).

So the two solutions are:

Why do we use \(\pm\)? Because both \(2^2\) and \((-2)^2\) equal 4.

Worked Example 3: A harder equation

Solve:

Step 1: Move the constant term.

Step 2: Complete the square.

Half of \(-4\) is \(-2\), and \((-2)^2=4\). Add 4 to both sides:

Step 3: Take square roots.

Step 4: Solve for \(x\).

So the solutions are:

Sometimes the answers are whole numbers, and sometimes they include square roots. Both are okay.

What if the coefficient of \(x^2\) is not 1?

We can still complete the square, but first we factor the coefficient of \(x^2\) out of the first two terms.

Worked Example 4: Completing the square when the coefficient is not 1

Rewrite:

Step 1: Factor 2 out of the first two terms.

Step 2: Complete the square inside the brackets.

Half of 4 is 2, and \(2^2=4\).

It is important to subtract 8, not 4, because the 4 is inside brackets multiplied by 2.

So the completed square form is:

This form is very useful for graphing because it shows the parabola’s turning point. In this case, the quadratic is in the form:

Here, \(2(x+2)^2-5\) can be thought of as:

So the turning point is at \((-2,-5)\).

Geometric idea behind completing the square

The name comes from geometry. Imagine \(x^2\) as the area of a square with side length \(x\). The term \(bx\) can be thought of as extra rectangular pieces added to that square.

When we take half of the \(x\)-coefficient and place one part on each side, we can form a larger square. The little missing corner is exactly the number we must add to complete the square.

You do not need to draw this every time, but it helps explain why the method works.

Common mistakes to avoid

• Forgetting to move the constant first when solving an equation
• Not adding the same number to both sides of an equation
• Forgetting to square after halving
• Forgetting the \(\pm\) when taking square roots
• Making errors when the coefficient of \(x^2\) is not 1

For example, if you have \(x^2+12x\), do not add 12. You must take half of 12, which is 6, and then square it. So you add 36.

Quick step-by-step guide

1. If solving an equation, move the constant term to the other side.
2. Look at the coefficient of \(x\).
3. Take half of it.
4. Square that number.
5. Add that value to both sides, or add and subtract it in an expression.
6. Rewrite as a perfect square binomial.
7. If solving, take square roots and finish solving for \(x\).

Mini check

Try these in your head:

• To complete the square for \(x^2+14x\), add and subtract \(49\).
• To complete the square for \(x^2-8x\), add and subtract \(16\).
• \(x^2+2x+1=(x+1)^2\)
• \(x^2-12x+36=(x-6)^2\)

Summary

Completing the square means rewriting a quadratic so that part of it becomes a perfect square. The main idea is to take half of the coefficient of \(x\), then square it.

This method helps solve quadratic equations and rewrite functions in a form like:

That form is especially useful because it shows important graph features like the turning point.

Deriving the Quadratic Formula

When we solve a quadratic equation, we are solving an equation with a highest power of 2, such as \(x^2+5x+6=0\).

Some quadratic equations can be solved by factoring, but not all of them factor nicely. That is why the quadratic formula is so useful: it gives a way to solve any quadratic equation written in the form

where \(a\neq 0\).

In this lesson, you will learn where the quadratic formula comes from. We will derive it step by step using a method called completing the square.

The quadratic formula is:

Instead of just memorizing it, we will build it ourselves.

Step 1: Start with the general quadratic equation

We begin with

To complete the square, the coefficient of \(x^2\) should be 1. Right now it is \(a\), so we divide every term by \(a\).

Now move the constant term to the other side:

Step 2: Complete the square

To complete the square, take half of the coefficient of \(x\), then square it.

The coefficient of \(x\) is \(\frac{b}{a}\). Half of that is

Now square it:

Add this value to both sides of the equation:

The left side is now a perfect square trinomial. It factors as

So we have

Step 3: Combine the right side into one fraction

To combine the terms, write \(-\frac{c}{a}\) with denominator \(4a^2\):

Now substitute that in:

Step 4: Take the square root of both sides

When we take the square root, we must remember the plus or minus symbol.

Simplify the square root:

Step 5: Solve for \(x\)

Subtract \(\frac{b}{2a}\) from both sides:

Now combine the terms over one denominator:

And that is the quadratic formula.

Why the formula matters

• It works for any quadratic equation.
• It is especially useful when factoring is difficult or impossible.
• It comes directly from completing the square, so it is not just a random rule.

A closer look at \(b^2-4ac\)

The expression under the square root, \(b^2-4ac\), is very important. It tells us what kind of answers the equation will have.

• If \(b^2-4ac>0\), there are two different real solutions.
• If \(b^2-4ac=0\), there is one real solution.
• If \(b^2-4ac<0\), there are no real solutions.

In 9th Grade, we usually focus on whether the solutions are real or not real.

Worked Example 1: Check the derivation with a simple quadratic

Solve \(x^2+6x+5=0\) by completing the square, and compare with the quadratic formula.

First, complete the square:

Half of 6 is 3, and \(3^2=9\). Add 9 to both sides:

Take square roots:

So:

The solutions are

Now use the quadratic formula with \(a=1\), \(b=6\), and \(c=5\):

This gives:

The answers match, which shows the formula works.

Worked Example 2: Use the formula when \(a\neq 1\)

Solve \(2x^2+7x-4=0\).

Identify the values:

Substitute into the formula:

Now find both solutions:

So the solutions are

Worked Example 3: A quadratic with one real solution

Solve \(x^2-8x+16=0\).

Here, \(a=1\), \(b=-8\), and \(c=16\).

There is only one real solution:

This happened because \(b^2-4ac=0\).

Worked Example 4: A quadratic with no real solutions

Solve \(x^2+2x+5=0\).

Let \(a=1\), \(b=2\), and \(c=5\).

The value inside the square root is negative, so there are no real solutions.

Common mistakes to avoid

• Forgetting to divide by \(a\) before completing the square in the derivation.
• Forgetting the \(\pm\) when taking a square root.
• Using the wrong signs for \(b\) or \(c\) when substituting into the formula.
• Not putting parentheses around negative numbers, such as writing \(-8^2\) instead of \((-8)^2\).

How to remember the derivation idea

1. Start with \(ax^2+bx+c=0\).
2. Divide by \(a\).
3. Move the constant term.
4. Complete the square.
5. Take square roots.
6. Solve for \(x\).

If you remember those steps, the quadratic formula becomes much easier to understand.

Brief Summary

The quadratic formula is derived by starting with the general quadratic equation and using completing the square. This process turns

into

The formula works for every quadratic equation and helps when factoring does not work easily. Understanding the derivation also helps you remember the formula and use it correctly.

Discriminant Analysis helps us predict what kind of solutions a quadratic equation will have before we solve it completely.

A quadratic equation is usually written in the form

$$ax^2 + bx + c = 0$$

where a \neq 0.

In this lesson, you will learn how to use the expression

$$b^2 - 4ac$$

to tell whether a quadratic has:

• two real solutions,
• one real solution, or
• no real solutions (which means the solutions are complex).

This expression is called the discriminant.

The discriminant comes from the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The part under the square root, \(b^2 - 4ac\), is very important. The value under a square root tells us what kind of answer we can get.

If the number under the square root is positive, we can take the square root and get two different real answers. If it is zero, the square root is zero, so both answers are the same. If it is negative, we cannot get a real square root, so there are no real solutions.

Here is the key idea:

• If \(b^2 - 4ac > 0\), there are two different real solutions.
• If \(b^2 - 4ac = 0\), there is one real solution (a repeated root).
• If \(b^2 - 4ac < 0\), there are two complex solutions and no real solutions.

You can also connect this to the graph of a quadratic, which is a parabola.

• Two real solutions: the parabola crosses the x-axis at two points.
• One real solution: the parabola touches the x-axis at one point.
• No real solutions: the parabola does not touch the x-axis.

So the discriminant tells us about both the equation and the graph.

How to use discriminant analysis

1. Write the quadratic in the form \(ax^2 + bx + c = 0\).
2. Identify the values of \(a\), \(b\), and \(c\).
3. Substitute them into \(b^2 - 4ac\).
4. Decide whether the result is positive, zero, or negative.
5. Use that result to describe the number and type of solutions.

Lets work through some examples.

Worked Example 1: Positive discriminant

Determine the number of solutions for

$$x^2 - 5x + 6 = 0$$

First, identify the coefficients:

• \(a = 1\)
• \(b = -5\)
• \(c = 6\)

Now calculate the discriminant:

$$b^2 - 4ac = (-5)^2 - 4(1)(6)$$

$$= 25 - 24$$

$$= 1$$

Since \(1 > 0\), the equation has two different real solutions.

That means the parabola crosses the x-axis twice.

Worked Example 2: Zero discriminant

Determine the number of solutions for

$$x^2 - 4x + 4 = 0$$

Identify the coefficients:

• \(a = 1\)
• \(b = -4\)
• \(c = 4\)

Now calculate:

$$b^2 - 4ac = (-4)^2 - 4(1)(4)$$

$$= 16 - 16$$

$$= 0$$

Since the discriminant is \(0\), the equation has one real solution.

This is sometimes called a double root because the same solution happens twice.

On the graph, the parabola touches the x-axis at exactly one point.

Worked Example 3: Negative discriminant

Determine the number of solutions for

$$2x^2 + 3x + 5 = 0$$

Identify the coefficients:

• \(a = 2\)
• \(b = 3\)
• \(c = 5\)

Now calculate:

$$b^2 - 4ac = 3^2 - 4(2)(5)$$

$$= 9 - 40$$

$$= -31$$

Since \(-31 < 0\), the equation has no real solutions.

It has two complex solutions, but in 9th Grade the main idea is that there are no real x-intercepts.

So the parabola does not cross or touch the x-axis.

Worked Example 4: Watch the signs carefully

Determine the number of solutions for

$$-3x^2 + 2x + 1 = 0$$

Identify the coefficients:

• \(a = -3\)
• \(b = 2\)
• \(c = 1\)

Now calculate:

$$b^2 - 4ac = 2^2 - 4(-3)(1)$$

$$= 4 - (-12)$$

$$= 4 + 12$$

$$= 16$$

Since \(16 > 0\), the equation has two different real solutions.

This example shows why signs matter. Because \(a\) was negative, the product \(4ac\) was negative, and subtracting a negative became addition.

Important reminders

• Always make sure the equation is in the form \(ax^2 + bx + c = 0\) before finding \(a\), \(b\), and \(c\).
• Be careful with negative numbers, especially when squaring \(b\).
• Remember that \((-5)^2 = 25\), not \(-25\).
• The discriminant tells the type of solutions, not necessarily the exact solutions.

Common mistakes

• Forgetting to set the equation equal to zero
  Example: for \(x^2 + 2x = 3\), rewrite it as \(x^2 + 2x - 3 = 0\).
• Using the wrong value for \(b\)
  In \(x^2 - 7x + 10 = 0\), the value of \(b\) is \(-7\), not \(7\).
• Not squaring the whole number
  \((-4)^2 = 16\).
• Mixing up the meaning of the result
  A negative discriminant means no real solutions, not one solution.

Quick check table

$$\begin{array}{|c|c|c|} \hline \text{Discriminant } (b^2 - 4ac) & \text{Type of solutions} & \text{Graph} \\ \hline > 0 & \text{Two different real solutions} & \text{Crosses x-axis twice} \\ \hline = 0 & \text{One real solution} & \text{Touches x-axis once} \\ \hline < 0 & \text{Two complex solutions, no real solutions} & \text{Does not touch x-axis} \\ \hline \end{array}$$

Practice thinking

Without fully solving, decide what each equation has:

• \(x^2 + x - 2 = 0\)
• \(4x^2 - 4x + 1 = 0\)
• \(3x^2 + 2x + 7 = 0\)

You would only need to calculate \(b^2 - 4ac\) for each one.

Brief answers:

• \(x^2 + x - 2 = 0\): \(1^2 - 4(1)(-2) = 1 + 8 = 9\), so two real solutions.
• \(4x^2 - 4x + 1 = 0\): \((-4)^2 - 4(4)(1) = 16 - 16 = 0\), so one real solution.
• \(3x^2 + 2x + 7 = 0\): \(2^2 - 4(3)(7) = 4 - 84 = -80\), so no real solutions.

Summary

The discriminant is the expression \(b^2 - 4ac\) from the quadratic formula. It helps you decide what kind of solutions a quadratic equation has without solving the whole equation.

If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. If it is negative, there are no real solutions.

When using discriminant analysis, always first write the equation in standard form, identify \(a\), \(b\), and \(c\), and then calculate carefully.

Modeling Projectile Motion helps us use quadratic functions to describe the path of an object that is thrown, kicked, or launched through the air.

When a ball is tossed upward and forward, its path usually forms a parabola. Because parabolas are modeled by quadratic functions, projectile motion is a real-life example of how quadratics are used.

In this lesson, you will learn how to:

• recognize when a quadratic function models a projectile,
• write and read an equation for height,
• find the maximum height,
• find how long the object stays in the air, and
• interpret the meaning of the graph.

1. What is projectile motion?

Projectile motion is the motion of an object moving through the air after it is thrown or launched.

Examples include:

• a basketball shot toward a hoop,
• a soccer ball kicked into the air,
• a rock tossed upward,
• a firework launched upward.

If we graph the object’s height against time, the graph is usually a parabola that opens downward.

That means the equation for height is often a quadratic function such as

$$h(t)=at^2+bt+c$$

Here:

• \(h(t)\) is the height of the object,
• \(t\) is the time,
• \(a\), \(b\), and \(c\) are numbers.

For projectile motion, the value of \(a\) is usually negative, so the parabola opens downward. This makes sense because the object rises, reaches a highest point, and then falls.

2. Understanding the parts of the equation

Suppose the height of an object is modeled by

$$h(t)=-5t^2+20t+3$$

Each part gives information:

• \(-5t^2\) makes the graph curve downward.
• \(20t\) affects how quickly the object rises at the start.
• \(+3\) is the starting height.

To find the starting height, plug in \(t=0\):

$$h(0)=-5(0)^2+20(0)+3=3$$

So the object starts 3 units above the ground.

3. The graph of projectile motion

A projectile’s graph has some important features:

• the y-intercept, which is the starting height,
• the vertex, which is the maximum height,
• the x-intercept(s), which show when the object is on the ground.

For projectile motion, the most important point is often the vertex, because it tells us the highest point reached.

4. Finding the maximum height

For a quadratic function

$$h(t)=at^2+bt+c$$

the time when the projectile reaches its maximum height is

$$t=\frac{-b}{2a}$$

After finding that time, substitute it back into the equation to get the maximum height.

This works because the vertex of a parabola occurs at

$$x=\frac{-b}{2a}$$

In projectile motion, the input variable is usually time, so we write

$$t=\frac{-b}{2a}$$

5. Finding how long the object stays in the air

The object hits the ground when its height is 0.

So to find the flight time, set

$$h(t)=0$$

Then solve the quadratic equation.

You might solve by:

• factoring,
• using the quadratic formula,
• or using a graph.

If you get two solutions, one may be negative. A negative time does not make sense in this situation, so use the positive solution.

Worked Example 1: Finding the starting height and whether the graph opens up or down

A toy rocket’s height is modeled by

$$h(t)=-4t^2+12t+2$$

Find:

1. the starting height,
2. whether the parabola opens up or down.

Step 1: Find the starting height.

The starting height happens at \(t=0\).

$$h(0)=-4(0)^2+12(0)+2=2$$

So the rocket starts at 2 units above the ground.

Step 2: Decide how the graph opens.

The coefficient of \(t^2\) is \(-4\), which is negative.

So the parabola opens downward.

Answer:

• Starting height: 2 units
• The graph opens downward

Worked Example 2: Finding the maximum height

A ball’s height is modeled by

$$h(t)=-5t^2+20t+1$$

Find the maximum height.

Step 1: Find the time of the maximum height.

Use

$$t=\frac{-b}{2a}$$

Here, \(a=-5\) and \(b=20\).

$$t=\frac{-20}{2(-5)}=\frac{-20}{-10}=2$$

So the ball reaches its maximum height at 2 seconds.

Step 2: Find the height at \(t=2\).

$$h(2)=-5(2)^2+20(2)+1$$

$$h(2)=-5(4)+40+1$$

$$h(2)=-20+40+1=21$$

So the maximum height is 21 units.

Answer: The ball reaches a maximum height of 21 units after 2 seconds.

Worked Example 3: Finding how long the object stays in the air

A ball is launched from the ground, and its height is modeled by

$$h(t)=-5t^2+15t$$

How long does it stay in the air?

Step 1: Set the height equal to 0.

$$-5t^2+15t=0$$

Step 2: Factor.

$$-5t(t-3)=0$$

Step 3: Solve.

Set each factor equal to 0:

$$-5t=0 \quad \text{or} \quad t-3=0$$

$$t=0 \quad \text{or} \quad t=3$$

The solution \(t=0\) is when the ball starts on the ground.

The solution \(t=3\) is when the ball returns to the ground.

Answer: The ball stays in the air for 3 seconds.

Worked Example 4: Finding both maximum height and flight time

A water balloon is thrown from a platform. Its height is modeled by

$$h(t)=-5t^2+30t+10$$

Find:

1. the maximum height,
2. how long it takes to hit the ground.

Part A: Maximum height

First find the time of the vertex:

$$t=\frac{-b}{2a}=\frac{-30}{2(-5)}=\frac{-30}{-10}=3$$

Now find the height at \(t=3\):

$$h(3)=-5(3)^2+30(3)+10$$

$$h(3)=-5(9)+90+10$$

$$h(3)=-45+90+10=55$$

So the maximum height is 55 units.

Part B: Time to hit the ground

Set the height equal to 0:

$$-5t^2+30t+10=0$$

To make this easier, divide both sides by \(-5\):

$$t^2-6t-2=0$$

This does not factor nicely, so use the quadratic formula:

$$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Here, \(a=1\), \(b=-6\), and \(c=-2\).

$$t=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-2)}}{2(1)}$$

$$t=\frac{6\pm\sqrt{36+8}}{2}$$

$$t=\frac{6\pm\sqrt{44}}{2}$$

Since \(\sqrt{44}\approx 6.63\),

$$t=\frac{6\pm 6.63}{2}$$

This gives two solutions:

$$t\approx \frac{12.63}{2}=6.315$$

$$t\approx \frac{-0.63}{2}=-0.315$$

The negative time does not make sense here, so ignore it.

Answer:

• Maximum height: 55 units
• Time to hit the ground: about 6.3 seconds

6. What the values mean in context

In projectile problems, it is very important to interpret the numbers correctly.

• If \(h(0)=c\), then \(c\) is the starting height.
• If the vertex is \((2, 21)\), then after 2 seconds the object reaches a maximum height of 21 units.
• If the positive x-intercept is \(3\), then the object hits the ground after 3 seconds.

Always check the units in the problem. The height might be in feet or meters, and time might be in seconds.

7. Common mistakes to avoid

• Forgetting to use \(h(t)=0\) when finding when the object hits the ground.
• Using a negative time as the final answer. Negative time usually does not make sense in these problems.
• Mixing up time and height. The input is time, and the output is height.
• Stopping after finding \(t=\frac{-b}{2a}\). That gives the time of the maximum height, not the height itself.

8. Quick problem-solving steps

When you see a projectile motion equation, follow these steps:

1. Identify the equation for height.
2. Find the starting height by using \(t=0\).
3. Find the maximum height using \(t=\frac{-b}{2a}\), then substitute back in.
4. Find flight time by setting \(h(t)=0\) and solving.
5. Check whether your answer makes sense in the real situation.

9. Brief summary

Projectile motion is often modeled by a quadratic function because the path of the object forms a parabola.

In an equation like \(h(t)=at^2+bt+c\), the starting height is found by plugging in \(t=0\), the maximum height comes from the vertex, and the flight time is found by solving \(h(t)=0\).

If you can read the meaning of the vertex and intercepts, you can understand important information about the motion of the object.