Topics
7. Quadratic Functions
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United States of AmericaFL
Grade 912

7.06 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 yo-yo's position with respect to time is a quadratic relation.

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

Example 3

The size of a snowball rolling down a hill can be modelled 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

MA.912.AR.3.4

Write a quadratic function to represent the relationship between two quantities from a graph, a written description or a table of values within a mathematical or real-world context.

MA.912.AR.3.6

Given an expression or equation representing a quadratic function, determine the vertex and zeros and interpret them in terms of a real-world context.

MA.912.AR.3.7

Given a table, equation or written description of a quadratic function, graph that function, and determine and interpret its key features.

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

MA.912.F.1.2

Given a function represented in function notation, evaluate the function for an input in its domain. For a real-world context, interpret the output.

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