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2. Quadratic Functions
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United States of AmericaTennessee
Integrated Math 2

2.06 Modeling with quadratic functions

Lesson
Practice
Lesson

Concept summary

The different quadratic forms are useful for modeling different quadratic scenarios based on what information is given.

The factored form of a quadratic function is useful when we have two points that can be represented as the x-intercepts.

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The x-intercepts often represent the points when a parabolic arc meets the ground.

For example, the feet of an arch can be thought of as x-intercepts when modeling the arch as a quadratic function.

The vertex form of a quadratic function is useful when we know the maximum or minimum point of a quadratic situation.

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For the vertex, the x-coordinate is where or when the maximum occurs and the y-coordinate is the minimum or maximum value.

For example, the point where a basketball shot reaches its peak can be thought of as the vertex, with the player being set as the origin.

The standard form of a quadratic function is more useful in algebraic scenarios when we know that we are combining a constant, some variable, and the square of that variable together, or when we are comparing different types of functions.

Worked examples

Example 1

The sum of two whole numbers is 24. Let one of the numbers be x.

a

Let y represent the product of the numbers. Form an equation for y in terms of x.

Approach

If we let one of the integers be represented by x, then the other integer must be 24-x.

Solution

The equation modeling the product of the two integers is:

y=x(24-x)

b

Find the largest possible value for the product.

Approach

Notice that our model is a parabola that opens down, the largest product will be the maximum value of y and will occur at the vertex.

Solution

To find the vertex, we first need to find the axis of symmetry. We can rewrite the model equation in factored form by taking a factor of -1 from the term (24-x). This gives us:

y=-x(x-24)

This tells us that the x-intercepts of the model are \left( 0,0 \right) and \left( 24,0 \right).

This means the axis of symmetry is the vertical line x=12, so the maximum value for y is when x=12. Substituting this into the model equation:

\displaystyle y\displaystyle =\displaystyle -x(x-24)Model equation
\displaystyle y\displaystyle =\displaystyle -12(12-24)Substitute in x=12
\displaystyle y\displaystyle =\displaystyle 144Simplify

So the largest possible value for the product is 144.

Example 2

Neva's yo-yo has a string length of 2\text{ ft}. When using the yo-yo, it takes 1 second for the yo-yo to go as low as possible. Assume that the yo-yo's position with respect to time is a quadratic relation.

a

Write an equation modeling the path of the yo-yo.

Approach

If we let time and height be our two variables (t,h), we can set t=0 to be when Neva releases the yo-yo and h=0 to be at the height of Neva's yo-yoing hand in feet. When the yo-yo reaches its lowest point, we can say that it is at the vertex \left( 1,-2 \right). Since we have vertex coordinates for our model, we want to use vertex form.

Solution

Substituting the coordinates of the vertex into the vertex form of a quadratic function gives us:

y=a(x-1)^2-2

for some value of a.

We know also that the yo-yo starts at the origin, so we can substitute the coordinates of the origin into the model equation to find a.

\displaystyle y\displaystyle =\displaystyle a(x-1)^2-2Model equation
\displaystyle 0\displaystyle =\displaystyle a(0-1)^2-2Substitute in the origin
\displaystyle 0\displaystyle =\displaystyle a-2Simplify
\displaystyle 2\displaystyle =\displaystyle aAdd 2 to both sides

We can model the path of the yo-yo with the quadratic equation:

y=2(x-1)^2-2

b

Create a graph to model the yo-yo's height against time and use it to predict the times when the yo-yo is 1 \text{ ft} from Neva's hand.

Approach

We can model the scenario with a graph by using the equation found in the previous part. We can then use the graph to find the times when the yo-yo is 1\text{ ft} from Neva's hand, which is when y=-1 since y is in feet and the yo-yo will be below Neva's hand.

Solution

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Using the model, we can see that y=-1 at approximately x=0.3 and x=1.7 seconds.

So we can predict that the yo-yo is 1\text{ ft} from Neva's hand 0.3 and 1.7 seconds after releasing the yo-yo.

Reflection

To check if our predictions are reasonable, we can substitute our predicted times into the equation and find the heights at those times.

\displaystyle y\displaystyle =\displaystyle 2(x-1)^2-2Model equation
\displaystyle y\displaystyle =\displaystyle 2(0.3-1)^2-2Substitute in x=0.3
\displaystyle y\displaystyle =\displaystyle -1.02Evaluate the expression
\displaystyle y\displaystyle =\displaystyle 2(x-1)^2-2Model equation
\displaystyle y\displaystyle =\displaystyle 2(1.7-1)^2-2Substitute in x=1.7
\displaystyle y\displaystyle =\displaystyle -1.02Evaluate the expression

This shows that our predictions are reasonable, since the height at our predicted times is very close to y=-1.

Example 3

The size of a snowball rolling down a hill can be modeled by the equation:

S=\frac{1}{2}t^2+2t+7

where S is the diameter of the snowball in inches and t is the time in seconds after the snowball starts rolling.

a

Find the value of S when t=4 and interpret the result in the given context.

Solution

To find S when t=4:

\displaystyle S\displaystyle =\displaystyle \frac{1}{2}t^2+2t+7Model equation
\displaystyle S\displaystyle =\displaystyle \frac{1}{2}\cdot 4^2+2\cdot 4+7Substitute in t=4
\displaystyle S\displaystyle =\displaystyle 23Simplify

This means that the diameter of the snowball will be 23 inches after the snowball has rolled down the hill for 4 seconds.

b

Identify the initial diameter of the snowball.

Approach

The snowball will be at its initial diameter when t=0.

Solution

\displaystyle S\displaystyle =\displaystyle \frac{1}{2}t^2+2t+7Model equation
\displaystyle S\displaystyle =\displaystyle \frac{1}{2}\cdot 0^2+2\cdot 0+7Substitute in t=0
\displaystyle S\displaystyle =\displaystyle 7Simplify

The initial diameter of the snowball was 7 inches.

Reflection

In standard form, the constant term is the y-value of the y-intercept.

Outcomes

M2.N.Q.A.1

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

M2.N.Q.A.1.A

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

M2.N.Q.A.1.C

Define and justify appropriate quantities within a context for the purpose of modeling.

M2.N.Q.A.1.D

Choose an appropriate level of accuracy when reporting quantities.

M2.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.*

M2.A.CED.A.3

Rearrange formulas to isolate a quantity of interest using algebraic reasoning.*

M2.F.IF.A.1

Use function notation.*

M2.F.IF.A.1.A

Use function notation to evaluate functions for inputs in their domains, including functions of two variables.

M2.F.IF.A.1.B

Interpret statements that use function notation in terms of a context.

M2.F.IF.B.4

Relate the domain of a function to its graph and, where applicable, to the context of the function it models. *

M2.MP1

Make sense of problems and persevere in solving them.

M2.MP2

Reason abstractly and quantitatively.

M2.MP3

Construct viable arguments and critique the reasoning of others.

M2.MP4

Model with mathematics.

M2.MP5

Use appropriate tools strategically.

M2.MP6

Attend to precision.

M2.MP7

Look for and make use of structure.

M2.MP8

Look for and express regularity in repeated reasoning.

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