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Algebra 1 View details

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Algebra 1 /

Chapter 9

Inverses of Quadratic Functions

Understanding the inverse of quadratic functions is pivotal in algebra. The lesson delves into the relationship between quadratic and square root functions, highlighting their ability to undo each other. A significant feature of a quadratic function is its rule involving raising the input value to the second power. To ensure a function's invertibility, it must pass the Horizontal Line Test. However, due to the symmetry of quadratic functions, inputs equidistant from the vertex get the same output. This symmetry necessitates domain restrictions to ensure unique outputs. The inverse operation of squaring is taking its square root, and this interplay between squaring and square root operations is a central theme. The lesson also provides practical examples, like determining a rocket's trajectory, to illustrate the application of these mathematical concepts.

Lesson Settings & Tools

	11 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 11

There are many applications of quadratic functions in real life. For example, the height of a projectile changes according to the time squared. A model using a quadratic function receives a time value

t

and returns the height

h

of the projectile at that time.

What if a person wants to know the time it takes the projectile to reach a specific height? The function could be used to set up a quadratic equation to find that time. However, if many predictions were needed, it would not be practical to solve multiple equations. It would be very convenient to have a function that takes the height and returns the corresponding time.

That convenient function is the inverse of a quadratic function. This lesson will begin by describing the relation between quadratic and square root functions. Then, how and why these functions can define inverses for each other will be understood.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Inverse Operation
Domain
Range
Inverse Relation
Quadratic Function
Vertex Form of a Quadratic Function
Square Root Function

Explore

Are Quadratic and Square Root Functions Related?

The following applet shows two graphs, one representing the quadratic function

f (x) = x^{2}

and the other is the square root function

g (x) = x .

In the applet, both functions have their domains defined as values of

x

greater than or equal to zero.

Interactive graph showing quadratic and square root functions

Can one graph be derived from the other? If so, how? Based on this exploration, is there any particular relationship between quadratic and square root functions?

Discussion

Inverse Functions

Every function has an inverse relation. If this inverse relation is also a function, then it is called an inverse function. In other words, the inverse of a function $f$ is another function $f^{- 1}$ such that they undo each other.

$f (f^{- 1} (x)) = x and f^{- 1} (f (x)) = x$

Also, if $x$ is the input of a function $f$ and $y$ its corresponding output, then $y$ is the input of $f^{- 1}$ and $x$ its corresponding output.

$f (x) = y \Leftrightarrow f^{- 1} (y) = x$

Example

Consider a function

f

and its inverse

f^{- 1} .

f (x) = 2 x - 3 and f^{- 1} (x) = \frac{x + 3}{2}

These functions will be shown to undo each other. To do so, it needs to be proven that

f (f^{- 1} (x)) = x

and that

f^{- 1} (f (x)) = x .

To start, the first equality will be proven. First, the definition of

f

will be used.

f (f^{- 1} (x)) = ? x ⇕ 2 f^{- 1} (x) - 3 = ? x

Now, in the above equation,

\frac{x + 3}{2}

will be substituted for

f^{- 1} (x) .

2 f^{- 1} (x) - 3 = ? x

Substitute

$f^{- 1} (x) = \frac{x + 3}{2}$

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

▼

Simplify left-hand side

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

(x + 3) - 3 = ? x

RemovePar

Remove parentheses

x + 3 - 3 = ? x

SubTerm

Subtract term

x = x ✓

A similar procedure can be performed to show that

f^{- 1} (f (x)) = x .

	Definition of First Function	Substitute Second Function	Simplify
$f (f^{- 1} (x)) = ? x$	$2 f^{- 1} (x) - 3 = ? x$	$2 (\frac{x + 3}{2}) - 3 = ? x$	$x = x ✓$
$f^{- 1} (f (x)) = ? x$	$\frac{f ( x ) + 3}{2} = ? x$	$\frac{2 x - 3 + 3}{2} = ? x$	$x = x ✓$

Therefore, $f$ and $f^{- 1}$ undo each other. The graphs of these functions are each other's reflection across the line $y = x .$ This means that the points on the graph of $f^{- 1}$ are the reversed points on the graph of $f .$

$Graph of the linear function f(x)=2*x-3 with a point (a,b) on it and the graph of its inverse function f^{-1}(x)=x/2+3/2 with a point (b,a) on it are depicted. The line of symmetry x=y for these functions is also shown.$

The characteristic feature of a quadratic function is that the function rule involves raising the input value to the second power. This feature is especially clear for the quadratic parent function.

f (x) = x^{2}

The inverse operation of raising a value to the second power is taking its square root. In other words, squaring a number and calculating its square root are operations that undo each other. The following example demonstrates this concept.

8^{2} = 6464 = 8

For this reason, square root functions can be used to define the inverse of quadratic functions. There is something else to consider, however. There is another possibility for the previous example.

(- 8)^{2} = 6464 = 8

The fact that the square of two different numbers can have the same value complicates defining an inverse for a quadratic function. In this case, the existence of an inverse operation is not enough for a function to be invertible. Before exploring the conditions a function must meet to be invertible, the definition of an invertible function will be given.

Concept

Invertible Function

A function is invertible if its inverse relation is also a function. For example, consider the following linear function.

f (x) = 0.5 x - 2

The graph of its inverse relation can be found by reflecting the graph of

f (x)

across the line

y = x .

Graph of the linear function f(x) =0.5x-2 and its inverse relation g(x)= 2x+4, which is also a linear function

The graph shows that the inverse relation,

g (x) = 2 x + 4,

is also a linear function. Therefore,

g (x) = f^{- 1} (x)

and, by definition,

f (x) = 0.5 x - 2

is invertible.

Discussion

Condition for Invertibility of a Function

There is a condition that a function must satisfy in order to be invertible.

For a function to be invertible, it must assign a unique output to every input.

This condition implies that if a function fails the Horizontal Line Test, then it is not invertible. However, some functions do not satisfy this condition for invertibility but can become invertible by restricting their domain. For example, consider the quadratic parent function $f (x) = x^{2} .$

This function has different $x -$ values where the same $y -$ value is assigned. The foll how the function fails the Horizontal Line Test.

The function f(x)=x^2 fails the Horizontal Line Test

Now, the graph of $f (x)$ will be reflected across the line $y = x$ to visualize its inverse relation.

Graph of f(x)=x^2 and its reflection across the line y=x

The inverse is not a function because it has input values with two different outputs assigned. The following graph further confirms this relation by showing how it fails the Vertical Line Test.

The relation plus / minus square root of x fails the Vertical Line Test

Therefore, $f (x) = x^{2}$ is not invertible because its inverse is not a function. Now, consider if the domain of $f (x)$ is restricted to $x \geq 0 .$ In that case, every $x -$ value gets assigned a unique $y -$ value. With the domain of $f (x)$ defined this way, $f^{- 1} (x) = x$ is a valid inverse function.

With the restricted domain an inverse function can be defined. This is illustrated showing the graphs of f(x)= x^2 and its inverse, square root of x, both of them defined for x equal or greater than 0

Proof

Consider a function $f (x)$ that assigns the same output to multiple inputs.

The corresponding inverse relation is not a function because it has multiple outputs assigned to the same input, which goes against the definition of function.

Therefore, to be invertible, it is a necessary condition that the function assigns a unique output to every input.

As previously shown, for a function to be invertible it must pass the Horizontal Line Test. Nevertheless, the graph of every quadratic function is a parabola with an axis of symmetry passing through its vertex. That means the inputs with the same distance from the vertex are assigned the same output.

The interactive graph shows that a parabola assigns the same output to the inputs with the same distance from the vertex

To avoid repeated output values and ensure invertibility, the domain needs to be restricted to values either equal or lower than or equal or greater than the

x -

coordinate of the vertex.

The function f(x)=x^2 can be divided into two pieces where the function assigns a unique output to every input

Example

Restricting the Domain of a Quadratic Function to Guarantee Invertibility

Heichi is so fascinated about the invertibility of quadratic functions, that he started a study group.

He brought the following function rule.

f (x) = \frac{1}{2} (x - 2)^{2} + 8

Heichi asks his friends to help him identify in which intervals the function is invertible. Select all the options that correspond to these intervals.

Hint

For a function to be invertible it must assign a unique output to every input.

Solution

It is a necessary condition for invertibility that the function assigns a unique output to every input. That being said, because of the symmetry of quadratic functions, inputs with the same distance from the vertex are assigned the same output.

To avoid having repeated output values, the domain needs to be restricted to values either equal or less than or equal or greater than the

x -

coordinate of the vertex. Since the given quadratic function is in vertex form, the vertex will be located by comparing the function with the general form.

Quatratic Function f (x) = a (x - h)^{2} + k Quatratic Function f (x) = \frac{1}{2} (x - 2)^{2} + 8 Vertex (h, k) Vertex (2, 8)

Because the

x -

coordinate of the vertex is

2,

the following intervals ensure that the function satisfies the invertibility condition.

I . II . - \infty < x \leq 2 - I 2 \leq x < \infty

Three of the five given options are contained in these intervals.

$- \infty < x \leq - 2$
$- \infty < x \leq - 2$
$- I 2 \leq x < \infty$

The following graph shows the corresponding pieces of the quadratic function obtained by restricting the domain to these three options. Note that the function assigns unique outputs to every input in all of them and, consequently, can be inverted.

The remaining two options are intervals where the function does not satisfy the invertibility condition. This can be verified by using the Horizontal Line Test.

Pop Quiz

Identifying Invertible Functions

The following applet shows different quadratic functions. The representation of these functions alternates between showing their graph and stating their function rule. Determine whether or not the given function is invertible in the indicated domain.

The applet shows a quadratic function with its domain restricted, alternating between its graphic and algebraic representation.

Discussion

Finding the Inverse of a Quadratic Function

A function can be represented by a table of values, a graph, a mapping diagram, or a function rule, among other ways. Depending on how the function is presented, finding its inverse can be done in different ways. When a function rule is given, finding the inverse algebraically is advantageous. Consider the following example function.

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

There is a series of steps to follow in order to find the inverse function

f^{- 1} (x) .

Replace

f (x)

With

y

To begin, since

f (x) = y

describes the input-output relationship of the function, replace

f (x)

with

y

in the function rule.

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

Switch

x

and

y

Because the inverse of a function reverses

x

and

y,

the variables can be switched. Notice that every other piece in the function rule remains the same.

y = \frac{2 x - 1}{3} switch x = \frac{2 y - 1}{3}

Solve for

y

Solve the resulting equation from the previous step for

y .

This will involve using the inverse operations.

x = \frac{2 y - 1}{3}

▼

Solve for

y

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

3 x = 2 y - 1

AddEqn

$LHS + 1 = RHS + 1$

3 x + 1 = 2 y

DivEqn

$LHS / 2 = RHS / 2$

\frac{3 x + 1}{2} = y

RearrangeEqn

Rearrange equation

y = \frac{3 x + 1}{2}

Replace

y

With

f^{- 1} (x)

Just as

f (x) = y

shows the input-output relationship of

f,

so does

f^{- 1} (x) = y .

Therefore, replacing

y

with

f^{- 1} (x)

gives the rule for the inverse of

f .

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

Notice that in

f,

the input is multiplied by

2,

decreased by

1,

and divided by

3 .

From the rule of

f^{- 1},

it can be seen that

x

undergoes the inverse of these operations in the reverse order. Specifically,

x

is multiplied by

3,

increased by

1,

and divided by

2 .

Suppose that the inverse of a quadratic function will be found algebraically. In that case, it is best that the function is written in vertex form before switching

x

and

y .

Vertex Form y = a (x - h)^{2} + k Switch x and y x = a (y - h)^{2} + k

If the function rule is in vertex before switching

x

and

y,

the power can be isolated. Then, the operation of raising the base to the second power can be undone by taking the square root, allowing to solve for

y

and find the inverse.

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

▼

Solve for

y

SubEqn

$LHS - k = RHS - k$

x - k = a (y - h)^{2}

DivEqn

$LHS / a = RHS / a$

\frac{x - k}{a} = (y - h)^{2}

SqrtEqn

$LHS = RHS$

\pm \frac{x - k}{a} = y - h

AddEqn

$LHS + h = RHS + h$

\pm \frac{x - k}{a} + h = y

RearrangeEqn

Rearrange equation

y = \pm \frac{x - k}{a} + h

Rewrite

Rewrite $y$ as $f^{- 1} (x)$

f^{- 1} (x) = \pm \frac{x - k}{a} + h

Recall that the domain of the quadratic function must be restricted for the function to be invertible. The choice of the square root sign depends on which piece of the quadratic function is used. The domain of

f (x)

must be equal to the range of

f^{- 1} (x)

and the range of

f (x)

must be equal to the domain of

f^{- 1} (x) .

Example

Modeling Falling Objects

Heichi took a few members from his study group named Ignacio and Kevin to try something cool in their apartment building. In their physics class, they learned that the height of an object that falls from an initial height

h

can be modeled using a quadratic function. They then adapted the function using the height of their building —

45

meters.

f (x) = - 5 x^{2} + h ⇓ f (x) = - 5 x^{2} + 45

Here

x

is the time in seconds and

f (x)

represents the height of the falling object, in meters. To test this model, Heichi goes to the top of the building and prepares to drop a tennis ball. Meanwhile, Ignacio and Kevin look attentive through their respective windows holding a chronometer as they wait for Heichi's signal.

Heich, Ignacio, and Kevin get prepared for the experiment

a Graph the given quadratic function. Identify the time interval starting when Heichi drops the ball from the top of the building until the moment it hits the ground and restrict the domain to this interval. What is the domain and range of the restricted function?

b Does the domain restriction indicated in Part A make the given function invertible? Justify the answer.

c What should be the inverse function's domain and range? Find the inverse function.

d Ignacio's window is

30

meters above the ground, and Kevin's window is

20

meters above the ground. Use the inverse function found in Part C to find the

x -

values corresponding to when the ball is in front of Ignacio's and Kevin's windows, respectively. Round the times to one decimal place.

Answer

a Graph:

Graph of the quadratic function f(x) = - 5x^2 +45

Domain: $0 \leq x \leq 3$

Range:

0 \leq y \leq 45

b Is the Function Invertible? Yes
Explanation: See solution.

c Domain:

0 \leq x \leq 45

Range: $0 \leq y \leq 3$

Inverse Function:

f^{- 1} (x) = \frac{4 5 - x}{5}

d Time to Reach Ignacio's Window:

1.7 s

Time to Reach Kevin's Window:

2.2 s

Hint

a The domain is the set of all the possible input values of a function, while the range is the set of all possible output values.

b For a function to be invertible, it must assign a unique output to every input.

c The inverse of a function reverses inputs and outputs.

d The inverse function

f^{- 1} (x)

receives as input the height of the ball with respect to the ground and returns as output the corresponding time that the ball takes to reach said height.

Solution

a First, the graph of the given quadratic function will be drawn.

Graph of the quadratic function y(x) = - 5t^2 +45

The ball starts falling at

x = 0

from a height of

45

meters. This corresponds to the vertex of the graph

(0, 45) .

On the other hand, the graph intersects the

x -

axis at

(3, 0) .

This means that the ball reaches the ground after

3

seconds. Note that

- 3

is another

x -

intercept but it is discarded because it represents a negative time.

Now, the function will be restricted to the time interval where the ball starts falling until it reaches the ground. The following graph represents the resulting piece of the quadratic function after being restricted.

The graph shows that the domain of the restricted function is

0 \leq x \leq 3

and the range is

0 \leq y \leq 45 .

b For a function to be invertible, it must assign a unique output value to every input. When considering the domain to be all real numbers, the function fails the Horizontal Line Test. This means that there are repeating outputs and, consequently, the function is not invertible.

However, all the $x -$ values in the interval $0 \leq x \leq 3$ have unique $y -$ values associated. This is why restricting the function's domain to the interval $0 \leq x \leq 3$ makes the function invertible.

c First the inverse will be found algebraically. Then, it will be explained what its domain and range must be and why. Since

f (x) = y

describes the input-output relationship of the function,

f (x)

will be replaced with

y

in the function rule.

f (x) = - 5 x^{2} + 45 \to y = - 5 x^{2} + 45

Now, the variables

y

and

x

will be switched because the inverse of a function reverses inputs and outputs.

y = - 5 x^{2} + 45 switch x = - 5 y^{2} + 45

Next, the resulting equation will be solved for

y .

x = - 5 y^{2} + 45

▼

Solve for

y

SubEqn

$LHS - 45 = RHS - 45$

x - 45 = - 5 y^{2}

DivEqn

$LHS / (- 5) = RHS / (- 5)$

\frac{x - 4 5}{- 5} = y^{2}

MoveNegDenomToNum

Put minus sign in numerator

\frac{- ( x - 4 5 )}{5} = y^{2}

Distr

Distribute $- 1$

\frac{- x + 4 5}{5} = y^{2}

CommutativePropAdd

Commutative Property of Addition

\frac{4 5 - x}{5} = y^{2}

SqrtEqn

$LHS = RHS$

\pm \frac{4 5 - x}{5} = y

RearrangeEqn

Rearrange equation

y = \pm \frac{4 5 - x}{5}

Finally,

y

will be replaced with

f^{- 1} (x) .

y = \pm \frac{4 5 - x}{5} ↓ f^{- 1} (x) = \pm \frac{4 5 - x}{5}

Recall that inverse functions are reflectionally symmetric with respect to

y = x .

The domain of

f^{- 1} (x)

must be equal to the range of

f (x),

which is

0 \leq y \leq 45 .

And the range

f^{- 1} (x)

must be equal to the domain of

f (x),

which is

0 \leq x \leq 3 .

Domain of f^{- 1} (x) : Range of f^{- 1} (x) : 0 \leq x \leq 45 0 \leq y \leq 3

Choosing the negative sign for the square root allows

f^{- 1} (x)

to assign negative outputs. Therefore,

f^{- 1} (x)

will only have the domain and range stated above if the positive sign is chosen.

f^{- 1} (x) = \frac{4 5 - x}{5}

Graphing the restricted quadratic function and the found inverse together shows how their graphs are reflections of each other across the line

y = x .

This confirms that the found inverse function is correct.

d To find the time it takes the ball to reach Igancio's window, which is at a height of

30

meters, the inverse function will be evaluated for

x = 30 .

f^{- 1} (x) = \frac{4 5 - x}{5}

▼

Substitute

30

for

x

and evaluate

Substitute

$x = 30$

f^{- 1} (x) = \frac{4 5 - 3 0}{5}

SubTerm

Subtract term

f^{- 1} (x) = \frac{1 5}{5}

CalcQuot

Calculate quotient

f^{- 1} (x) = 3

CalcRoot

Calculate root

f^{- 1} (x) = 1.732050 \dots

RoundDec

Round to $1$ decimal place(s)

f^{- 1} (x) = 1.7

Therefore, the ball will reach Ignacio's window

1.7

seconds after being dropped. Similarly, since Kevin's window is at a height of

20

meters, the time that the ball takes to reach it will be found by evaluating the inverse for

x = 20 .

f^{- 1} (x) = \frac{4 5 - x}{5}

▼

Substitute

20

for

x

and evaluate

Substitute

$x = 20$

f^{- 1} (x) = \frac{4 5 - 2 0}{5}

SubTerm

Subtract term

f^{- 1} (x) = \frac{2 5}{5}

CalcQuot

Calculate quotient

f^{- 1} (x) = 5

CalcRoot

Calculate root

f^{- 1} (x) = 2.236067 \dots

RoundDec

Round to $1$ decimal place(s)

f^{- 1} (x) = 2.2

Hence, ball will reach Kevin's window after

2.2

seconds.

Pop Quiz

Practice Finding a Point on the Inverse Function

The following applet shows a quadratic function and a specific point on its graph. The domain of the given function is restricted and the function is invertible. Find the coordinates of the corresponding point on the graph of its inverse function. Write the answer rounding to two decimal places.

Interactive applet showing different quadratic functions and a point on their graph

Example

Analyzing and Predicting the Trajectory of a Projectile

Heichi, Ignacio, and Kevin want to try another experiment. They bought a rocket model and found a website that predicts a projectile's trajectory. After inputting their building's height and the launching angle on the website, the calculations indicated that the following quadratic function would describe the trajectory of the rocket.

f (x) = x - \frac{x ^{2}}{1 6 0} + 50

The variable

x

represents the horizontal displacement from the building while

f (x)

represents the height from the ground, both measured in meters. The kids want the rocket to land inside a bucket that they plan to place on the top of a wooden ladder. Help them do the calculations needed to succeed.

Heichi, Ignacio, and Kevin get ready for the rocket launching

a Write the given quadratic function in vertex form and choose the corresponding option.

b The domain needs to be restricted for the function to be invertible. Keeping in mind that

x

represents the horizontal displacement from the building and

f (x)

represents the height from the ground, which of the following options represents the appropriate domain restriction?

What is the range of the restricted function?

c What is the inverse function of

f (x)

in the domain specified in Part B?

d Use the inverse function to determine the horizontal distance from the building the ladder needs to be so the rocket lands inside the bucket on top of it. The ladder the kids are using is

3

meters tall. Give the distance rounded to two decimal places.

Hint

a Complete the square for the quadratic expression of the function rule. Then, rewrite the expression as the square of a binomial plus a constant term.

b The format for a quadratic function in vertex form is

f (x) = (x - h)^{2} + k,

where

h

and

k

are the

x -

and

y -

coordinates of the vertex, respectively.

c The inverse of a function inverts the original function's inputs and outputs.

d Evaluate the inverse function for

x = 3

to find the horizontal distance of the rocket when its height is

3

meters.

Solution

a Complete the square of the quadratic expression involved to write the function in vertex form. The first step in doing so is to factor out the coefficient of the

x^{2} -

term.

f (x) = x - \frac{x ^{2}}{1 6 0} + 50 ⇕ f (x) = - \frac{1}{1 6 0} (x^{2} - 160 x - 8000)

The resulting expression inside parentheses has a linear coefficient of

b = - 160 .

Since

(\frac{b}{2})^{2}

is the addend needed to obtain a perfect square trinomial, this quantity will be calculated next.

(\frac{b}{2})^{2}

▼

Substitute

- 160

for

b

and evaluate

Substitute

$b = - 160$

(\frac{- 1 6 0}{2})^{2}

CalcQuot

Calculate quotient

(- 80)^{2}

CalcPow

Calculate power

6400

Therefore, by adding and subtracting

(\frac{b}{2})^{2} = 6400

x^{2} - 160 x - 8000,

a perfect square trinomial and an additional constant term will be obtained. The result will then be rewritten as the square of a binomial plus a constant term.

x^{2} - 160 x - 8000

▼

Rewrite

AddDiffZero

$a = a + 6400 - 6400$

x^{2} - 160 x - 8000 + 6400 - 6400

CommutativePropAdd

Commutative Property of Addition

x^{2} - 160 x + 6400 - 6400 - 8000

SubTerm

Subtract term

x^{2} - 160 x + 6400 - 14400

$a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2}$

(x - 80)^{2} - 14400

Finally, the result found above will be used to rewrite the given function in vertex form.

f (x) = - \frac{1}{1 6 0} (x^{2} - 160 x - 8000)

▼

Rewrite

Rewrite $x^{2} - 160 x - 8000$ as $(x - 80)^{2} - 14400$

f (x) = - \frac{1}{1 6 0} ((x - 80)^{2} - 14400)

Distr

Distribute $- \frac{1}{1 6 0}$

f (x) = - \frac{1}{1 6 0} (x - 80)^{2} + \frac{1 4 4 0 0}{1 6 0}

CalcQuot

Calculate quotient

f (x) = - \frac{1}{1 6 0} (x - 80)^{2} + 90

b A quadratic function has to have its domain restricted to be invertible. And, to ensure invertibility, the coordinates of the vertex must be known. Since the function is already in vertex form, it will be compared to the general format of the vertex form to identify the vertex location.

Quatratic Function f (x) = a (x - h)^{2} + k Quatratic Function f (x) = - \frac{1}{1 6 0} (x - 80)^{2} + 90 Vertex (h, k) Vertex (80, 90)

Note that

x

represents horizontal displacement and

f (x)

represents height. That means in this situation they both only make sense if they are non-negative.

x \geq 0 and f (x) \geq 0

The following graph of

f (x) = x - \frac{x ^{2}}{1 6 0} + 50

is shown bounded with these restrictions.

The given quadratic function with restricted domain

This graph shows the whole trajectory of the rocket if it were to land on the ground. Note that this function does not pass the Horizontal Line Test, so it is not invertible. The domain has to be further restricted. If the values of

x

greater than or equal to

80

and less than or equal to

200

are considered, the function will become invertible.

The graph shows the domain and the range of the restricted quadratic function.

Domain : Range : 80 \leq x \leq 200 0 \leq y \leq 90

This graph represents the part of the trajectory after the projectile reaches its highest point and starts falling. Remember that the kids want the rocket to land in a bucket placed on a ladder. That makes this part of the trajectory very useful.

c To invert the function, first,

f (x)

will be replaced with

y

in the function rule.

f (x) = - \frac{1}{1 6 0} (x - 80)^{2} + 90 ↓ y = - \frac{1}{1 6 0} (x - 80)^{2} + 90

Next, the variables

y

and

x

will be switched because the inverse of a function reverses inputs and outputs.

y = - \frac{1}{1 6 0} (x - 80)^{2} + 90 ↓ switch x = - \frac{1}{1 6 0} (y - 80)^{2} + 90

The resulting equation will be solved for

y .

x = - \frac{1}{1 6 0} (y - 80)^{2} + 90

▼

Solve for

y

SubEqn

$LHS - 90 = RHS - 90$

x - 90 = - \frac{1}{1 6 0} (y - 80)^{2}

MultEqn

$LHS \cdot (- 160) = RHS \cdot (- 160)$

- 160 (x - 90) = (y - 80)^{2}

Distr

Distribute $- 160$

- 160 x + 14400 = (y - 80)^{2}

CommutativePropAdd

Commutative Property of Addition

14400 - 160 x = (y - 80)^{2}

SqrtEqn

$LHS = RHS$

14400 - 160 x = y - 80

AddEqn

$LHS + 80 = RHS + 80$

\pm 14400 - 160 x = y - 80

RearrangeEqn

Rearrange equation

y = \pm 14400 - 160 x + 80

Finally,

y

will be replaced with

f^{- 1} (x) .

y = \pm 14400 - 160 x + 80 ↓ f^{- 1} (x) = \pm 14400 - 160 x + 80

Because the inverse function reverses inputs and outputs, the domain of

f^{- 1} (x)

must be equal to the range of

f (x) .

Similarly, the range

f^{- 1} (x)

must be equal to the domain of

f (x) .

Considering this relationship, the domain and range of

f^{- 1} (x)

can be stated.

Domain of f^{- 1} (x) : Range of f^{- 1} (x) : 0 \leq x \leq 90 0 \leq y \leq 200

Choosing the negative sign for the square root allows

f^{- 1} (x)

to assign negative outputs. Therefore,

f^{- 1} (x)

will only have the domain and range stated above if the positive sign is chosen.

f^{- 1} (x) = 14400 - 160 x + 80

d To find the horizontal distance of the rocket when its height is

3

meters, the inverse function will be evaluated for

x = 3 .

f^{- 1} (x) = 14400 - 160 x + 80

Substitute

$x = 3$

f^{- 1} (3) = 14400 - 160 (3) + 80

▼

Evaluate right-hand side

Multiply

f^{- 1} (3) = 14400 - 480 + 80

SubTerm

Subtract term

f^{- 1} (3) = 13920 + 80

CalcRoot

Calculate root

f^{- 1} (3) = 117.983049 \dots + 80

AddTerms

Add terms

f^{- 1} (3) = 197.983049 \dots

RoundDec

Round to $2$ decimal place(s)

f^{- 1} (3) \approx 197.98

For the rocket to land inside the bucket, the boys should place the ladder with the bucket

197.98

meters away from their building.

Closure

The Relation Between Quadratic and Radical Functions

At the beginning of this lesson, a graph showing a quadratic and a square root function was presented. Both functions had their domains defined as

0 \leq x .

Using the information from this lesson, it can be stated that these functions are in fact, inverse functions because

f (x) = x^{2}

and

g (x) = x

are reflections of each other across the line

y = x .

Interactive graph showing quadratic and radical functions

Additionally, depending on how the domain of function

f (x)

is defined, it can have two different inverse functions.

Domain of $f (x)$	Inverse Function
$0 \leq x < \infty$	$f^{- 1} (x) = x$
$- \infty < x \leq 0$	$f^{- 1} (x) = - x$

A great method to explore this is by using the following interactive graph. The graph gives various choices of domains for $f (x) = x^{2} .$ The function's corresponding inverse relation is shown by reflecting $f (x)$ across $y = x .$

Graph of f(x)=x^2 and its reflection across the line y=x for different domain definitions

Inverses of Quadratic Functions

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