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11.6: Applications with quadratic functions

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    45115
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    There are many applications involving quadratic functions that it is almost challenging to pick the few functions to discuss for this section. Yet, we only choose a few applications that are common in most algebra classes. First, we must start with defining the extreme value of a quadratic function.

    Find the Extreme Value

    Definition: Extreme Value

    The extreme value of a quadratic function \(f(x)\) is either the minimum or maximum value of the quadratic function \(f(x)\). The extreme value is given as

    \[f\left(-\dfrac{b}{2a}\right)\nonumber\]

    The minimum value is located at the lowest point of an upwards parabola and the maximum value is located at the highest point of a downwards parabola.

    Note

    Notice in the definition that the extreme value is the \(y\)-coordinate of the vertex. Recall, the vertex is

    \[\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)\nonumber\]

    and we can see the \(y\)-coordinate is just the extreme value. Hence, when we use the words vertex, minimum or maximum value, or extreme value, they are all associated with the vertex of a parabola.

    Example 11.6.1

    Find the vertex and the extreme value of the function \(f(x) = x^2 − 4x + 5\).

    Solution

    Notice \(a = 1\), which means \(a > 0\). Hence, \(f(x)\) is an upwards parabola and, from the definition, we expect \(f(x)\) to have a minimum value. Let’s use the formula to find the vertex, where \(a = 1\) and \(b = −4\).

    \[\begin{array}{rl}x=-\dfrac{b}{2a}&\text{Plug-n-chug} \\ x=-\dfrac{\color{blue}{-4}}{\color{blue}{2(1)}}&\text{Simplify} \\ x=2&\text{The }x\text{-coordinate of the vertex}\end{array}\nonumber\]

    Next, we find the \(y\)-coordinate of the vertex by obtaining \(f(2)\).

    \[\begin{array}{rl}f(x)=x^2-4x+5&\text{Plug-n-chug} \\ f(\color{blue}{2}\color{black}{)}=(\color{blue}{2}\color{black}{)}^2-4(\color{blue}{2}\color{black}{)}+5&\text{Evaluate} \\ f(2)=1&\text{The }y\text{-coordinate of the vertex}\end{array}\nonumber\]

    Hence, the vertex is \((2, 1)\). Next, we find the extreme value of \(f(x)\). From the vertex calculation, we see \(f\left(-\dfrac{b}{2a}\right)=f(2)=1\). Thus, the extreme value is \(1\), the \(y\)-coordinate of the vertex.

    Example 11.6.2

    Find the vertex and the extreme value of the function \(q(n) = −3n^2 − 5n + 3\).

    Solution

    Notice \(a = −3\), which means \(a < 0\). Hence, \(q(n)\) is an downwards parabola and, from the definition, we expect \(q(n)\) to have a maximum value. Let’s use the formula to find the vertex, where \(a = −3\) and \(b = −5\).

    \[\begin{array}{rl}n=-\dfrac{b}{2a}&\text{Plug-n-chug} \\ n=-\dfrac{\color{blue}{-5}}{\color{blue}{2(-3)}}&\color{black}{\text{Simplify}} \\ n=-\dfrac{5}{6}&\text{The }x\text{-coordinate of the vertex}\end{array}\nonumber\]

    Next, we find the \(y\)-coordinate of the vertex by obtaining \(q\left(-\dfrac{5}{6}\right)\).

    \[\begin{array}{rl} q(n)=-3n^2-5n+3&\text{Plug-n-chug} \\ q\left(\color{blue}{-\dfrac{5}{6}}\color{black}{}\right)=-3\left(\color{blue}{-\dfrac{5}{6}}\color{black}{}\right)^2-5\left(\color{blue}{-\dfrac{5}{6}}\color{black}{}\right) +3&\text{Evaluate} \\ q\left(-\dfrac{5}{6}\right)=\dfrac{61}{12}&\text{The }y\text{-coordinate of the vertex}\end{array}\nonumber\]

    Hence, the vertex is \(\left(-\dfrac{5}{6},\dfrac{61}{12}\right)\). Next, we find the extreme value of \(q(n)\). From the vertex calculation, we see \(q\left(-\dfrac{b}{2a}\right)=q\left(-\dfrac{5}{6}\right)=\dfrac{61}{12}\). Thus, the extreme value is \(\dfrac{61}{12}\), the \(y\)-coordinate of the vertex.

    It may seem a little redundant to find the vertex and the extreme value, but, remember, the goal is to apply this concept to real-world applications with quadratic functions. Let’s take a look at a few applications where we find the extreme value in context of a real-world model.

    Projectile Motion

    Example 11.6.3

    A rocket is launched at \(t = 0\) seconds. Its height, in meters above sea-level, as a function of time is given by \(h(t) = −4.9t^2 + 46t+ 157\). At what time does the rocket reach its maximum height? At what height does the rocket reach its maximum height above the water? Round the answers to 2 decimal places.

    Solution

    When we read the word maximum, we should think about the vertex of \(h(t)\). Since we need to find the time in which the maximum height occurs, then we can find the \(x\)-coordinate of the vertex.

    \[\begin{aligned}t&=-\dfrac{b}{2a} \\ t&=-\dfrac{46}{2(-4.9)} \\ t&\approx 4.69\end{aligned}\]

    Thus, the maximum height occurs after \(4.69\) seconds. Next, we find the height of the rocket when it reaches its maximum height above the water. Since we need to find the maximum height, then we need to find the \(y\)-coordinate of the vertex, or \(h(4.69)\).

    \[h(4.69)=-4.9(4.69)^2+46(4.69)+157\approx 264.96\nonumber\]

    Thus, the maximum height of the rocket is \(264.96\) meters after \(4.69\) seconds the rocket is launched.

    Revenue and Cost Functions

    Example 11.6.4

    The revenue, \(R(x)\), of producing and selling \(x\) Awesome Hearing Aids is modeled by the function \(R(x) = −6x^2 + 108x\). What is the maximum revenue?

    Solution

    To find the maximum revenue, we need to find \(R\left(-\dfrac{b}{2a}\right)\), where \(a = −6\) and \(b = 108\). Let’s plug-n-chug this into \(R(x)\) to find the maximum revenue.

    \[\begin{aligned} R(x)&=-6x^2+108x \\ R\left(-\dfrac{b}{2a}\right)&=-6\left(-\dfrac{b}{2a}\right)^2+108\left(-\dfrac{b}{2a}\right) \\ R\left(-\dfrac{\color{blue}{108}}{2\color{blue}{(-6)}}\right)&=-6\left(-\dfrac{\color{blue}{108}}{2\color{blue}{(-6)}}\right)^2+108\left(-\dfrac{\color{blue}{108}}{2\color{blue}{(-6)}}\right) \\ R\left(-\dfrac{\color{blue}{108}}{2\color{blue}{(-6)}}\right)&=486\end{aligned}\]

    Thus, the maximum revenus is $486.

    Example 11.6.5

    The cost, \(C(x)\), of producing \(x\) Totally Cool Coolers is modeled by the function \(C(x) = 0.005x^2 −0.3x+ 17\). What is the minimum cost?

    Solution

    To find the minimum cost, we need to find \(C\left(-\dfrac{b}{2a}\right)\), where \(a=0.005\) and \(b=-0.3\). Let's plug-n-chug this into \(C(x)\) to find the minimum cost.

    \[\begin{aligned}C(x)&=0.005x^2-0.3x+17 \\ C\left(-\dfrac{b}{2a}\right)&=0.005\left(-\dfrac{b}{2a}\right)^2-0.3\left(-\dfrac{b}{2a}\right)+17 \\ C\left(-\dfrac{\color{blue}{-0.3}}{2\color{blue}{(0.005)}}\right)&=0.005\left(-\dfrac{\color{blue}{-0.3}}{2\color{blue}{(0.005)}}\right)^2-0.3\left(-\dfrac{\color{blue}{-0.3}}{2\color{blue}{(0.005)}}\right)+17 \\ C\left(-\dfrac{\color{blue}{-0.3}}{2\color{blue}{(0.005)}}\right)&=12.5\end{aligned}\]

    Note

    From all the examples, we see the variety of methods in obtaining the extreme value of a quadratic function. We can either graph the function, find each coordinate of the vertex, or directly calculate the extreme value. It is at the discretion of the student to use any method. However, directly calculating the extreme value is recommended when only the extreme value is needed.

    Applications with Quadratic Functions Homework

    Exercise 11.6.1

    Find the vertex and the extreme value of the function \[f(x)=2x^2-5x-4\nonumber\] What is the vertex? What is the extreme value?

    Exercise 11.6.2

    Find the vertex and the extreme value of the function \[f(x)=2x^2-2x-5\nonumber\] What is the vertex? What is the extreme value?

    Exercise 11.6.3

    Find the vertex and the extreme value of the function \[f(x)=x^2+2x+24\nonumber\] What is the vertex? What is the extreme value?

    Exercise 11.6.4

    A rocket is launched at \(t = 0\) seconds. Its height, in meters above sea-level, as a function of time is given by \[h(t)=-4.9t^2+271t+150\nonumber\] At what time does the rocket reach its maximum height? At what height does the rocket reach its maximum height above the water? Round your answer to 2 decimal places.

    Exercise 11.6.5

    A rocket is launched at \(t = 0\) seconds. Its height, in meters above sea-level, as a function of time is given by \[h(t)=-4.9t^2+190t+395\nonumber\] At what time does the rocket reach its maximum height? At what height does the rocket reach its maximum height above the water? Round your answer to 2 decimal places.

    Exercise 11.6.6

    A rocket is launched at \(t = 0\) seconds. Its height, in meters above sea-level, as a function of time is given by \[h(t)=-4.9t^2+223t+129\nonumber\] At what time does the rocket reach its maximum height? At what height does the rocket reach its maximum height above the water? Round your answer to 2 decimal places.

    Exercise 11.6.7

    The cost, \(C(x)\), of producing \(x\) Totally Cool Coolers is modeled by the function \[C(x)=0.005x^2-0.25x+12\nonumber\] How many coolers need to be produced and sold in order to minimize the cost? What is the cost?

    Exercise 11.6.8

    The cost, \(C(x)\), of producing \(x\) Totally Cool Coolers is modeled by the function \[C(x)=0.005x^2-0.45x+25\nonumber\] How many coolers need to be produced and sold in order to minimize the cost? What is the cost?

    Exercise 11.6.9

    The revenue, \(R(x)\), of producing and selling \(x\) Awesome Hearing Aids is modeled by the function \[R(x)=-5x^2+105x\nonumber\] How many hearing aids need to be produced and sold in order to maximize the revenue? What is the revenue?

    Exercise 11.6.10

    The revenue, \(R(x)\), of producing and selling \(x\) Awesome Hearing Aids is modeled by the function \[R(x)=-2x^2+62x\nonumber\] How many hearing aids need to be produced and sold in order to maximize the revenue? What is the revenue?

    Exercise 11.6.11

    The revenue, \(R(x)\), of producing and selling \(x\) Awesome Hearing Aids is modeled by the function \[R(x)=-4x^2+76x\nonumber\] How many hearing aids need to be produced and sold in order to maximize the revenue? What is the revenue


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