Topics
7. Quadratic Functions
topic badge
United States of AmericaTennessee
Algebra 1

7.03 Quadratic functions in factored form

Lesson
Practice
Lesson

Concept summary

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

\displaystyle y=a(x-x_1)(x-x_2)
\bm{x_1, \,x_2}
x-values of the x-intercepts
\bm{a}
The scale factor, which tells us about the shape of the graph
-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y
If a>0, then the quadratic function opens upwards and has a minimum value.
-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y
If a<0, then the quadratic function opens downwards and has a maximum value.

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

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. As the graph has a line of symmetry passing through the vertex, we know the vertex lies half way 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.

Worked examples

Example 1

A quadratic function in factored form has the equation:

y=2(x-4)(x+6)

a

State the coordinates of the x-intercepts.

Approach

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.

Solution

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

Reflection

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

b

Determine the coordinates of the y-intercept.

Approach

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.

Solution

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 in x=0
\displaystyle y\displaystyle =\displaystyle 2(-4)(6)Simplify the factors
\displaystyle y\displaystyle =\displaystyle -48Simplify

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

c

Determine the coordinates of the vertex.

Approach

The vertex lies on the axis of symmetry, so the x-coordinate of the vertex will be the average of 4 and -6. We can then substitute this value into the funtion to find the y-coordinate.

Solution

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 in x=-1
\displaystyle y\displaystyle =\displaystyle 2(-5)(5)Simplify the factors
\displaystyle y\displaystyle =\displaystyle -50Simplify

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

d

Draw the graph of the function.

Approach

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.

Solution

-8
-6
-4
-2
2
4
6
8
x
-60
-50
-40
-30
-20
-10
10
20
y

Reflection

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 2

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

The path of the cannon ball 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.

Approach

To make the graph match the information, we want to make sure that our choice of axis labels makes sense for the context, and the axis scales correctly describe the path of the cannon ball. Both axes will have meters as their units.

Solution

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 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 cannon ball starts at the edge of the cliff at \left(0,15\right), reaches it peak at \left(10,20\right), and then falls into the sea at \left(30,0\right).

5
10
15
20
25
30
x\left(\text{m}\right)
5
10
15
20
y\left(\text{m}\right)

Reflection

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 cannon ball.

Approach

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 cannon ball, 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 relation, 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.

Solution

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 in \left(0,15\right)
\displaystyle 15\displaystyle =\displaystyle a(10)(-30)Simplify the parentheses
\displaystyle 15\displaystyle =\displaystyle -300aEvaluate the product
\displaystyle -\frac{1}{20}\displaystyle =\displaystyle aDivide both sides by -300

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

c

A second cannon ball 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 cannon ball landed.

Approach

To find where the cannon ball landed, we'll find the point where y is 0 since the y-value represents height above sea level, and a height of 0 indicates sea level. We'll need to find the corresponding x-value.

Solution

\displaystyle y\displaystyle =\displaystyle -\frac{1}{15}(x+12)(x-27)Model equation
\displaystyle 0\displaystyle =\displaystyle -\frac{1}{15}(x+12)(x-27)Substitute in y=0
\displaystyle 0\displaystyle =\displaystyle (x+12)(x-27)Divide both sides by -\frac{1}{15}
\displaystyle x\displaystyle =\displaystyle -12,x=27Solve for x

Since the cannon ball 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 cannon ball lands in the sea 27 meters from the base of the cliff.

Reflection

Since the model equation was in factored form, we could also have identified the x-intercepts directly from the equation.

Outcomes

A1.N.Q.A.1

Use units as a way to understand real-world problems.*

A1.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.*

A1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

A1.A.REI.D.5

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

A1.F.IF.C.7

Graph functions expressed algebraically and show key features of the graph by hand and using technology.*

A1.F.IF.C.8

Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. *

A1.F.IF.C.8.A

Rewrite quadratic functions to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a real-world context.

A1.F.IF.C.9

Compare properties of functions represented algebraically, graphically, numerically in tables, or by verbal descriptions.*

A1.F.IF.C.9.A

Compare properties of two different functions. Functions may be of different types and/or represented in different ways.

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP2

Reason abstractly and quantitatively.

A1.MP3

Construct viable arguments and critique the reasoning of others.

A1.MP4

Model with mathematics.

A1.MP6

Attend to precision.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

What is Mathspace

About Mathspace