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7. Quadratic Functions
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Algebra 1

7.02 Quadratic functions in factored form

Lesson
Practice

Quadratic functions in factored form

Exploration

Consider the graph below, which shows the vertical position, y in feet, of a water balloon thrown by a child from the low diving board of a pool over time, x in seconds:

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The function representing the projectile motion of the water balloon is y=-3x^2+3x+6.

  1. Shorena says that the function y=-3(x+1)(x-2) is equivalent to the given function. How can we determine if she is correct?
  2. How do the characteristics of the graph relate to the context?
  3. How might the function y=-3(x+1)(x-2) relate to the graph?

One way to represent quadratic functions is using the factored form. This form allows us to identify the x-intercepts, the direction of opening, and scale factor of the quadratic function.

\displaystyle f(x)=a(x-x_1)(x-x_2)
\bm{x_1, \,x_2}
x-values of the x-intercepts
\bm{a}
scale factor
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If a>0, then the quadratic function opens upwards and has a minimum value.
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If a<0 then the quadratic function opens downwards and has a maximum value.

The x-intercepts are the points where f(x)=0, so we refer to x_1 and x_2 as the zeros of the function.

x
y
  • (x_1,\,0) and (x_2,\,0) are the x-intercepts of the function y=f(x)

  • x_1 and x_2 are zeros of the function

  • (x - x_1) and (x - x_2) are factors of the function y=f(x)

  • x_1 and x_2 are solutions or roots of the equation f(x)=0

To draw the graph of a quadratic function, we generally want to find three different points on the graph, such as the x- and y-intercepts.

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Since the graph of a quadratic function has a line of symmetry passing through the vertex, we know the vertex lies halfway between the two x-intercepts.

We can also determine the direction in which the graph opens by identifying if the scale factor, a, is positive or negative.

Examples

Example 1

Consider the graph of a quadratic function:

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a

Identify the coordinates of the x- and y-intercepts of the function.

Worked Solution
Create a strategy

The x-intercepts occur when y=0 and the y-intercept occurs when x=0.

Apply the idea
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The x-intercepts are \left( -2,\,0 \right) and \left( 3,\,0 \right).

The y-intercept is \left( 0,\,2 \right).

b

Find the equation of the quadratic function in factored form.

Worked Solution
Create a strategy

Substitute the x-intercepts for x_1 and x_2 in the equation y=a(x-x_1)(x-x_2), then use any other point on the graph to substitute for x and y and solve for a.

Apply the idea

Since the x-values of the x-intercepts are -2 and 3, we know that the factored form will be:

y=a(x+2)(x-3)

for some value of a. We can find a by substituting in the coordinates of the y-intercept into the function.

To find a:

\displaystyle y\displaystyle =\displaystyle a(x+2)(x-3)Factored form
\displaystyle 2\displaystyle =\displaystyle a(0+2)(0-3)Substitute \left(0,2\right)
\displaystyle 2\displaystyle =\displaystyle a(2)(-3)Evaluate the addition and subtraction
\displaystyle 2\displaystyle =\displaystyle -6aEvaluate the multiplication
\displaystyle -\frac{1}{3}\displaystyle =\displaystyle aDivide both sides by -6

The equation of the quadratic function in factored form:

y=-\frac{1}{3}(x+2)(x-3)

Example 2

Consider the quadratic function:

y=2x^{2} + 4x - 48

a

State the coordinates of the x-intercepts.

Worked Solution
Create a strategy

In the factored form y=a(x-x_1)(x-x_2), the values of x_1 and x_2 are the x-values of the x-intercepts. The y-value of the x-intercepts is y=0. Factor the quadratic, then determine its x-intercepts.

We can factor out a GCF of 2, so that the equation becomes y=2(x^2+2x-24).

Since there are no common factors for the remaining three terms and the trinomial is not a perfect square trinomial, we proceed to factor by grouping by finding the value of two integers that multiply to ac = (1)(-24) = -24 and add up to b = 2. After finding these integers, we use them to rewrite the middle term 2x as a sum of two terms.

Apply the idea

The factor pair whose sum is 2 is -4 and 6.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle 2(x^2+2x-24)\displaystyle =\displaystyle 2(x^2-4x + 6x - 24)Rewrite polynomial with four terms
\displaystyle =\displaystyle 2[x(x-4) +6(x-4)]Factor each pair
\displaystyle =\displaystyle 2(x-4)(x+6)Divide out common factor of (x-4)

There are no more common factors to be divided out, so the fully factored form of the quadratic function is y=2(x-4)(x+6).

The x-intercepts are \left( 4,\,0 \right) and \left( -6,\,0 \right).

Reflect and check

Notice that x+6 is the same as x-(-6).

b

Determine the coordinates of the y-intercept.

Worked Solution
Create a strategy

The y-value of the y-intercept is the result when x=0. We can substitute x=0 into the factored form to find this value.

Apply the idea

To find the y-value of the y-intercept:

\displaystyle y\displaystyle =\displaystyle 2(x-4)(x+6)Given quadratic function
\displaystyle y\displaystyle =\displaystyle 2(0-4)(0+6)Substitute x=0
\displaystyle y\displaystyle =\displaystyle 2(-4)(6)Evaluate the subtraction and addition
\displaystyle y\displaystyle =\displaystyle -48Evaluate the multiplication

The y-intercept is \left( 0,\,-48 \right).

Reflect and check

The original function shows the y-intercept, which we can identify without making any calculations.

c

Determine the coordinates of the vertex.

Worked Solution
Create a strategy

The vertex lies on the axis of symmetry, so the x-coordinate of the vertex will be exactly in the middle between the two x-intercepts. We can find the middle value by taking the average of 4 and -6. We can then substitute this x-coordinate value into the function to find the y-coordinate.

Apply the idea

To find the x-coordinate:

The average of 4 and -6 is half way between them. We can calculate that \dfrac{4+(-6)}{2}=-1, so the x-coordinate of the vertex and the axis of symmetry is x=-1.

To find the y-coordinate:

\displaystyle y\displaystyle =\displaystyle 2(x-4)(x+6)Given quadratic function
\displaystyle y\displaystyle =\displaystyle 2(-1-4)(-1+6)Substitute x=-1
\displaystyle y\displaystyle =\displaystyle 2(-5)(5)Evaluate the subtraction and addition
\displaystyle y\displaystyle =\displaystyle -50Evaluate the mutliplication

The vertex is \left( -1,\,-50 \right).

d

Draw the graph of the function.

Worked Solution
Create a strategy

The scale factor is 2 which is positive, so the graph will open up. We can draw the graph through any three points that we know are on it.

Apply the idea
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Reflect and check

Any three points is enough to draw the graph, but knowing where the vertex is can make it easier since the vertex is on the axis of symmetry.

Example 3

Identify the characteristics of h\left(x\right)=\dfrac{1}{6}\left(3x + 2\right)\left(x - 7\right).

a

Identify the factors of the function.

Worked Solution
Create a strategy

The function is given in factored form y=a\left(x-x_1\right)\left(x-x_2\right) where a,\,\left(x-x_1\right) and \left(x-x_2\right) are the factors.

Apply the idea

The factors of h\left(x\right) are \dfrac{1}{6},\,\left(3x+2\right) and \left(x-7\right).

Reflect and check

\dfrac{1}{6} is the greatest common factor (GCF) of a\left(x-x_1\right)\left(x-x_2\right) but is still a factor of the function.

b

Identify the roots of the function.

Worked Solution
Create a strategy

The roots of the function are the x values, x_1 and x_2, where h\left(x\right)=0.

Apply the idea

To solve for the roots algebraically, set each of the variable factors equal to zero.3x+2=0 \text{ and } x-7=0

Rearrange each equation to isolate the term with the variable.3x=-2 \text{ and } x=7

Isolate the variable by dividing by the coefficient of x.x=-\dfrac{2}{3} \text{ and } x=7

The roots of h(x) are x=-\dfrac{2}{3},\,7

Reflect and check

Notice in part (a) we also identified the factor of \dfrac{1}{6} but we did not use it to find roots. That is because a factor without a variable will not result in a root because \dfrac{1}{6} \neq0.

We can substitute these values back into the equation to check our answers. If we substitute {x=-\dfrac{2}{3}} and x=7 and get h\left(x\right)=0, then we know our roots are correct.

Let's start with the root x=-\dfrac{2}{3}.

\displaystyle h(x)\displaystyle =\displaystyle \dfrac{1}{6}\left(3\left(-\dfrac{2}{3}\right) + 2\right)\left(-\dfrac{2}{3} - 7\right)Substitue x=-\dfrac{2}{3}
\displaystyle =\displaystyle \dfrac{1}{6}\left(-2 + 2\right)\left(\dfrac{2}{3} - 7\right)Evaluate the multiplication
\displaystyle =\displaystyle \dfrac{1}{6}\left(0\right)\left(-\dfrac{23}{3}\right)Evaluate inside the parentheses
\displaystyle =\displaystyle 0Zero product property

Next, let's try the root x=7.

\displaystyle h(x)\displaystyle =\displaystyle \dfrac{1}{6}\left(3\cdot 7 + 2\right)\left(7 - 7\right)Substitue x=7
\displaystyle =\displaystyle \dfrac{1}{6}\left(21 + 2\right)\left(7 - 7\right)Evaluate the multiplication
\displaystyle =\displaystyle \dfrac{1}{6}\left(23\right)\left(0\right)Evaluate inside the parentheses
\displaystyle =\displaystyle 0Zero product property

Evaluating for each root gave an output of 0 confirming that both are infact roots of the function.

c

Identify the zeros of the function.

Worked Solution
Create a strategy

The zeros of a function are the same as its roots.

Apply the idea

In part (b) we solved for the roots, x_1 and x_2, and got x=-\dfrac{2}{3},\, 7. These are also the zeros of the function.

d

State the x-intercepts of the function.

Worked Solution
Create a strategy

The points \left(x_1,\,0\right) and \left(x_2,\,0\right) are the x-intercepts for h\left(x\right).

Apply the idea

The zeros or roots of h\left(x\right) are -\dfrac{2}{3} and 7.

These are the x-values of the x-intercepts. The y-value of any y-intercept is 0 because the x-axis is at x=0.

The x-intercepts of the function are at \left(-\dfrac{2}{3},\,0\right) and \left(7,\,0\right).

Reflect and check

We can check our x-intercepts by graphing h\left(x\right).

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There are two points where the parabola crosses the x-axis, at x=-\dfrac{2}{3} and x=7.

Example 4

The graph of a quadratic function has x-intercepts at \left(-2,\,0\right) and \left(1,\,0\right) and passes through the point \left(-3,\,-2\right). Write an equation in factored form that models this quadratic.

Worked Solution
Create a strategy

To write the equation for this quadratic in factored form we need to first identify the roots or zeros of the equation. We can then substitute these values for x_1 and x_2.

The x-intercepts of the function are at \left(-2,\,0\right) and \left(1,\,0\right), so we know the equation has roots/zeros of {x=-2} and x=1.

Apply the idea

Since the zeros are x=-2 and x=1, we can identify the factors by rearranging those equations so they are equal to 0:

By adding 2 to both sides of x=-2 and subtracting 1 from both sides of x=1 we get: x+2=0 \text{ and } x-1=0

We can put these in the factored form as the factors:y=a(x+2)(x-1)

We can find a by substituting the coordinates of the additional point, \left(-3,\,-2\right), into the function.

To find a:

\displaystyle y\displaystyle =\displaystyle a(x+2)(x-1)Factored form
\displaystyle -2\displaystyle =\displaystyle a(-3+2)(-3-1)Substitute x=-3 and y=-2
\displaystyle -2\displaystyle =\displaystyle a(-1)(-4)Evaluate the addition
\displaystyle -2\displaystyle =\displaystyle 4aEvaluate the multiplication
\displaystyle -\dfrac{1}{2}\displaystyle =\displaystyle aDivide both sides by 4

Substituting the value we found for a, the equation of the quadratic function in factored form is:

y=-\dfrac{1}{2}(x+2)(x-1)

Reflect and check

Checking the graph of the equation, we can see that it satisfies the given information.

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Example 5

Find the equation that models the graph shown below.

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Worked Solution
Create a strategy

This quadratic function only has 1 x-intercept, which is also the vertex. When this happens, the function is in the form f(x)=a(x-x_1)^2. Remember, we need an additional point, like the y-intercept, to find the exact equation to this function.

Apply the idea

Since the x-intercept is at (-2,\,0), the function takes the form f(x)=a(x+2)^2.

Next, we can use the y-intercept of \left(0,\,-1\right) to find the value of the leading coefficient.

\displaystyle f(x)\displaystyle =\displaystyle a(x+2)^2Given equation
\displaystyle -1\displaystyle =\displaystyle a(0+2)^2Substitute x=0 and y=-1
\displaystyle -1\displaystyle =\displaystyle a(2)^2Evaluate the addition
\displaystyle -1\displaystyle =\displaystyle 4aEvaluate the exponent
\displaystyle -\dfrac{1}{4}\displaystyle =\displaystyle aDivide both sides by 4

The equation of the graph is f(x)=-\dfrac{1}{4}(x+2)^2.

Reflect and check

We could have used any point on the parabola to solve for the scale factor, a. There is another point at (-6,\,-4). We would substitute x=-6 and y=-4, then the equation would take the form -4=a(-6+2)^2.

\displaystyle -4\displaystyle =\displaystyle a(-6+2)^2
\displaystyle -4\displaystyle =\displaystyle a(-4)^2
\displaystyle -4\displaystyle =\displaystyle 16a
\displaystyle -\dfrac{1}{4}\displaystyle =\displaystyle a

Notice that this is the same thing we got earlier because no matter which points we substitute in we will get the same function because they are all points on the same parabola.

Example 6

A cannonball is fired from the edge of a cliff which is 15 meters above sea level. The peak of the cannonball's arc is 20 meters above sea level and 10 meters horizontally from the cliff edge. The cannonball lands in the sea 30 meters away from the base of the cliff.

The path of the cannonball is shown on the following graph, but the axes have not been labeled.

a

Label the axes of the graph to match the information provided.

Worked Solution
Create a strategy

To make the graph match the information, we want to make sure that the axes labels and scales accurately represent the path of the cannonball and make sense for the context. Both axes will have meters as their units.

Apply the idea

We can see that the path on the graph starts at a point on the vertical axis and ends at a point on the horizontal axis. So we can make the vertical axis represent the height, with y=0 being sea level, and the horizontal axis represent distance, with x=0 being the edge of the cliff.

We can then add values onto the axes to show that the cannonball starts at the edge of the cliff at \left(0,\,15\right), reaches its peak at \left(10,\,20\right), and then falls into the sea at \left(30,\,0\right).

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Reflect and check

Another way to show the scale of the axes is to label some key points. For example:

x\left(\text{m}\right)
y\left(\text{m}\right)
b

Determine the factored equation which models the path of the cannonball.

Worked Solution
Create a strategy

To match the graph in part (a), we can let x represent the horizontal distance from the cliff, and let y represent the height above sea level.

To find the factored equation that models the cannonball, we need to know both x-intercepts and the scale factor.

We know that one of the x-intercepts is at x=30, and that the vertex is at x=10. Remember that the vertex lies on the axis of symmetry of a quadratic function, so we can use this to find the other x-intercept.

We can find the scale factor by substituting any point into the equation (that isn't an x-intercept) and solving for the scale factor that makes the equation true.

Apply the idea

Since both x-intercepts have the same y-value, they will be mirrored across the axis of symmetry. Since x=30 is 20 more units than x=10, the other x-intercept will be at 20 less units than x=10. So the other x-intercept is at x=-10.

If we let the scale factor of the equation be a, then our equation will be: y=a\left(x+10\right)\left(x-30\right)

We can find the scale factor by substituting in a point on the graph (let's use the y-intercept) and solving for a.

\displaystyle y\displaystyle =\displaystyle a(x+10)(x-30)Model equation
\displaystyle 15\displaystyle =\displaystyle a(0+10)(0-30)Substitute \left(0,\,15\right)
\displaystyle 15\displaystyle =\displaystyle a(10)(-30)Evaluate the addition and subtraction
\displaystyle 15\displaystyle =\displaystyle -300aEvaluate the multiplication
\displaystyle -\frac{1}{20}\displaystyle =\displaystyle aDivide both sides by -300

So the equation which models the path of the cannonball is: y=-\frac{1}{20}\left(x+10\right)\left(x-30\right)

c

A second cannonball is fired, and this one can be modeled by the equation: y=-\frac{1}{15}\left(x+12\right)\left(x-27\right)Use this model to predict where the cannonball landed.

Worked Solution
Create a strategy

We can use what we know about the general factored form of the equation to provide information about the cannonball's path.

Apply the idea

Since the a value is negative, we know this function will open downward. This makes sense for a cannonball, as it will arc upwards to a maximum vertical height before falling back down, due to gravity. The x-intercepts will be at -12 and 27. Since the cannonball is being fired away from the cliff in the positive x-direction, we know that x=-12 is a non-viable solution. So the second cannonball lands in the sea 27 meters from the base of the cliff.

Idea summary

To write the equation of the graph of a quadratic function in factored form, substitute the x-intercepts for x_1 and x_2 in the equation y=a(x-x_1)(x-x_2), then use any other point on the graph to substitute for x and y and solve for a, the scale factor.

Outcomes

A.F.2

The student will investigate, analyze, and compare characteristics of functions, including quadratic and exponential functions, and model quadratic and exponential relationships.

A.F.2b

Given an equation or graph, determine key characteristics of a quadratic function including x-intercepts (zeros), y-intercept, vertex (maximum or minimum), and domain and range (including when restricted by context); interpret key characteristics as related to contextual situations, where applicable.

A.F.2c

Graph a quadratic function, f(x), in two variables using a variety of strategies, including transformations f(x) + k and kf(x), where k is limited to rational values.

A.F.2d

Make connections between the algebraic (standard and factored forms) and graphical representation of a quadratic function.

A.F.2g

For any value, x, in the domain of f, determine f(x) of a quadratic or exponential function. Determine x given any value f(x) in the range of f of a quadratic function. Explain the meaning of x and f(x) in context.

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