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Intermediate Algebra Functions and Graphs

Katherine Yoshiwara

Section A.4 Chapter 4 Applications of Quadratic Models

Subsection A.4.1 Quadratic Formula

Subsubsection A.4.1.1 Simplify square roots

Be careful when simplifying radicals.
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Example A.4.1.
Can you simplify the first expression into the second expression? (Decide whether the expressions are equivalent.)
  1. Is \(~~\sqrt{4+x^2}~~\) equivalent to \(~~2+x\text{?}\)
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  2. Is \(~~\sqrt{\dfrac{x^2}{9}}~~\) equivalent to \(~~\dfrac{x}{3}~~\) for \(~x \ge 0\text{?}\)
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  3. Is \(~~\sqrt{w-3}~~\) equivalent to \(~~\sqrt{w} - \sqrt{3}\text{?}\)
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Solution.
  1. If the expressions are equivalent, they must be equal for every value of the variable. Let’s test with \(x=3\text{.}\) Then
    \begin{align*} \sqrt{4+x^2} \amp = \sqrt{4+9} = \sqrt{13} \approx 3.6\\ \text{but}~~~~~~~~~ 2+x \amp = 2+3 = 5 \end{align*}
    No, the expressions are not equivalent.
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  2. Because \(\left(\dfrac{x}{3}\right)^2 = \dfrac{x}{3} \cdot \dfrac{x}{3} = \dfrac{x^2}{3^2} = \dfrac{x^2}{9}\text{,}\) it is also true that \(\sqrt{\dfrac{x^2}{9}} = \dfrac{x}{3}\text{.}\) Yes, the expressions are equivalent.
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  3. Let \(w=16\text{.}\) Then
    \begin{align*} \sqrt{w-3} \amp = \sqrt{16-3} = \sqrt{13} \approx 3.6\\ \text{but}~~~ \sqrt{w}-\sqrt{3} \amp = \sqrt{16}-\sqrt{3} \approx 4-1.7 = 2.3 \end{align*}
    No, the expressions are not equivalent.
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Decide whether the expressions are equivalent. Assume all variables are positive.
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Checkpoint A.4.3.
\(\sqrt{64x^2y^2}~\) and \(~8xy\)
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Answer.
Yes
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Checkpoint A.4.4.
\(\sqrt{64+x^2y^2}~\) and \(~8+xy\)
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Answer.
No
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Checkpoint A.4.5.
\(\sqrt{\dfrac{c^2+d^2}{4b^2}}~\) and \(~\dfrac{\sqrt{c^2+d^2}}{2b}\)
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Answer.
Yes
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Subsubsection A.4.1.2 Use the order of operations

In the order of operations, simplifying radicals comes after what’s inside parentheses (or fraction bars) and before products and quotients.
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Example A.4.6.
Simplify each expression. Do not use a calculator!
  1. \(\displaystyle \dfrac{4-\sqrt{64}}{2}\)
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  2. \(\displaystyle -2\left(3\sqrt{16}-\sqrt{3(27)}\right)\)
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  3. \(\displaystyle 6-3\sqrt[3]{27-7(5)}\)
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Solution.
  1. We start by simplifying the numerator.
    \begin{align*} \dfrac{4-\sqrt{64}}{2} \amp = \dfrac{4-8}{2} \amp\amp \blert{\text{Evaluate the radical, then subtract.}}\\ \amp = \dfrac{-4}{2}=-2 \amp\amp \blert{\text{Reduce the fraction.}} \end{align*}
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  2. We start by simplifyng what’s inside parentheses.
    \begin{align*} -2\amp\left(3\sqrt{16} -\sqrt{3(27)}\right) \amp\amp \blert{\text{Evaluate the radicals.}}\\ \amp = -2(3 \cdot 4-\sqrt{81}) \amp\amp \blert{\text{Simplify inside the parentheses.}}\\ \amp = -2(12-9)=-6 \end{align*}
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  3. We start by simplifying the radicand.
    \begin{align*} 6-3\sqrt[3]{27-7(5)} \amp = 6-3\sqrt[3]{27-35} \amp\amp \blert{\text{Subtract under the radical.}}\\ \amp = 6-3\sqrt[3]{-8} \amp\amp \blert{\text{Evaluate the radical.}}\\ \amp =6-3(-2)=12 \end{align*}
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Example A.4.7.
Simplify each expression. Round your answer to hundredths.
  1. \(\displaystyle \dfrac{8-2\sqrt{2}}{4}\)
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  2. \(\displaystyle 2+6\sqrt[3]{-25}\)
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Solution.
  1. Do not start with "\(8-2\)"! Evaluate \(\sqrt{2}\) first, then multiply by 2, and subtract the result from 8. Once the numerator is simplified, divide by 4.
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    On a calculator, enter
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    \(\qquad\qquad\) ( \(8\) - \(2~\boxed{\sqrt{}}\) ) ) ÷ \(4\) ENTER
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    and round to two decimal places: \(~1.29\)
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  2. Evaluate the cube root, multiply by 6, then add the result to 2.
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    On a calculator, enter
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    \(\qquad\qquad 2\) + \(6\) MATH 4 - \(25\) ENTER
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    and round to two decimal places: \(~{-15.54}\)
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Checkpoint A.4.8.
Simplify each expression. Do not use a calculator!
  1. \(\displaystyle \dfrac{36}{6+\sqrt{36}}\)
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  2. \(\displaystyle 10+2(3-\sqrt{169})\)
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  3. \(\displaystyle \dfrac{3+\sqrt[3]{-729}}{6-\sqrt[3]{-27}}\)
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Answer.
  1. \(\displaystyle 3\)
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  2. \(\displaystyle -10\)
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  3. \(\displaystyle \dfrac{-2}{3}\)
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Checkpoint A.4.9.
Simplify each expression. Round your answer to hundredths.
  1. \(\displaystyle \dfrac{6+9\sqrt{3}}{3}\)
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  2. \(\displaystyle -1-3\sqrt[3]{120}\)
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Answer.
  1. \(\displaystyle 7.20\)
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  2. \(\displaystyle -15.80\)
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Subsubsection A.4.1.3 Use the quadratic formula

Example A.4.10.
  1. Use the quadratic formula to solve the equation \(~x = \dfrac{2}{3} - \dfrac{x^2}{6}\)
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  2. Give decimal approximations to two decimal places for the solutions.
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Solution.
  1. To clear fractions, multiply each term by the LCD, \(6\text{,}\) to get
    \begin{align*} \alert{6}(x) \amp = \alert{6}(x) \left(\dfrac{2}{3} - \dfrac{x^2}{6}\right)\\ 6x \amp = 4 - x^2 \end{align*}
    Next, write the equation in standard form and identify the coefficients.
    \begin{gather*} x^2+6x-4 = 0\\ a = 1, b = 6, c = -4 \end{gather*}
    Substitute the coefficients into the quadratic formula and sinplify.
    \begin{align*} x \amp = \dfrac{-6 \pm \sqrt{6^2-4(1)(-4)}}{2(1)}\\ \amp = \dfrac{-6 \pm \sqrt{52}}{2} \end{align*}
    The solutions are \(\dfrac{-6 + \sqrt{52}}{2}\) and \(\dfrac{-6 - \sqrt{52}}{2}\)
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  2. We can use a calculator to find decimal approximations for the solutions.
    \begin{gather*} \dfrac{-6 + \sqrt{52}}{2} \approx \dfrac{-6 + 7.21}{2} = 0.605\\ \dfrac{-6 - \sqrt{52}}{2} \approx \dfrac{-6 - 7.21}{2} = -6.605 \end{gather*}
    To two decimal places, the solutions are \(0.61\) and \(-6.61\text{.}\)
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Checkpoint A.4.11.
Use the quadratic formula to solve the equation \(x^2 + 2x = 5\)
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Answer.
\(x = \dfrac{-2 \pm \sqrt{24}}{2}\)
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Subsection A.4.2 The Vertex

Subsubsection A.4.2.1 Parabolas in vertex form

We can sketch the graph of a parabola with the vertex, the \(y\)-intercept, and its symmetric point.
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Example A.4.12.
Graph the equation \(~y=2(x-15)^2-72\)
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Solution.
The vertex is \((15,-72)\text{.}\) We find the \(y\)-intercept by setting \(x=0\text{:}\)
\begin{equation*} y=2(\alert{0}-15)^2-72=378 \end{equation*}
The \(y\)-intercept is \((0, 378)\text{.}\) We find the \(x\)-intercepts by setting \(y=0\text{:}\)
\begin{align*} 2(x-15)^2-72 \amp = \alert{0}\\ (x-15)^2 \amp = 36\\ x \amp = \pm 6+15 \end{align*}
The \(x\)-intercepts are \((9,0)\) and \((21,0)\text{.}\) We plot these three points and sketch the parabola through them.
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parabola
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Checkpoint A.4.13.
Find the vertex, the \(y\)-intercept, and the \(x\)-intercepts of \(~y=-2(x+1)^2+3\text{,}\) and sketch its graph.
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Answer.
parabola
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Subsubsection A.4.2.2 Product of binomials

With practice, you will be able to multiply binomials mentally.
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Example A.4.14.
Multiply \(~(x-6)(x+3)\text{.}\)
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Solution.
We apply the distributive law to multiply each term of the first factor by each term of the second factor. (The "FOIL" method.)
\begin{align*} (x-6)(x+3) \amp = x \cdot x + 3 \cdot x - 6 \cdot x - 6 \cdot 3\\ \amp = x^2+3x-6x-18 \amp \amp \blert{\text{Combine like terms.}}\\ \amp = x^2-3x-18 \end{align*}
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Example A.4.15.
Expand \(~(2t-5)^2\text{.}\)
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Solution.
Multiply \(~(2t-5)(2t-5)\)
\begin{align*} (2t-5)(2t-5) \amp = 4t^2-10t-10t+25 \amp \amp \blert{\text{Combine like terms.}}\\ \amp = 4t^2-20t+25 \end{align*}
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Checkpoint A.4.16.
Multiply \(~(4a-7)(3a+8)\text{.}\)
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Answer.
\(12a^2+11a-56\)
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Checkpoint A.4.17.
Expand \(~(3-4b)^2\text{.}\)
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Answer.
\(16b^2-24b+9\)
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Subsubsection A.4.2.3 Solve for a parameter

We can find the equation for a parabola if we know, for example, the vertex and one other point.
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Example A.4.18.
The point \((6,2)\) lies on the graph of \(~y=a(x-4)^2+1.~\) Solve for \(a\text{.}\)
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Solution.
Substitute 6 for \(x\) and 2 for \(y\text{,}\) then solve for \(a\text{.}\)
\begin{align*} \alert{2} \amp = a(\alert{6}-4)^2+1\\ 2 \amp = a(4) + 1\\ 1 \amp = 4a \end{align*}
The solution is \(a = \dfrac{1}{4}\text{.}\)
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Example A.4.19.
The point \((-2,11)\) lies on the graph of \(~y=x^2+bx-3.~\) Solve for \(b\text{.}\)
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Solution.
Substitute -2 for \(x\) and 11 for \(y\text{,}\) then solve for \(b\text{.}\)
\begin{align*} (\alert{-2})^2 +b(\alert{-2}) -3 \amp = 11\\ 4-2b-3 \amp = 11\\ -2b \amp = 10 \end{align*}
The solution is \(b = -5\text{.}\)
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Checkpoint A.4.20.
The point \((-6,10)\) lies on the graph of \(~y=a(x+3)^2-2.~\) Solve for \(a\text{.}\)
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Answer.
\(a=\dfrac{4}{3}\)
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Checkpoint A.4.21.
The point \((-3,8)\) lies on the graph of \(~y=-x^2+bx+5.~\) Solve for \(b\text{.}\)
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Answer.
\(b=-4\)
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Checkpoint A.4.22.
The point \((8,-12)\) lies on the graph of \(~y=ax^2-4x+36.~\) Solve for \(a\text{.}\)
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Answer.
\(a=\dfrac{-1}{3}\)
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Checkpoint A.4.23.
The point \((60,-480)\) lies on the graph of \(~y=\dfrac{-2}{3}(x-h)^2+120.~\) Solve for \(h\text{.}\)
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Answer.
\(h=30\)
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Subsection A.4.3 Curve Fitting

Subsubsection A.4.3.1 Points on a graph

If a curve passes through a given point, the coordinates of the point satisfy the equation of the curve.
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Example A.4.24.
Write an equation to say that \((-3,8)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
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Solution.
Substitute \(-3\) for \(x\) and \(8\) for \(y\text{.}\)
\begin{align*} 8 \amp = a(-3)^2+b(-3)+c \amp \amp \blert{\text{Simplify.}}\\ 8 \amp = 9a-3b+c \end{align*}
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Checkpoint A.4.25.
Write an equation to say that \((-4,-18)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
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Answer.
\(-16a-4b+c=-18\)
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Checkpoint A.4.26.
Write an equation to say that \((8,0)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
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Answer.
\(64a+8b+c=0\)
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Checkpoint A.4.27.
Write an equation to say that \((0,-5)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
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Answer.
\(c=-5\)
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Checkpoint A.4.28.
Write an equation to say that \((-60,400)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
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Answer.
\(3600a-60b+c=400\)
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Subsubsection A.4.3.2 Solve a 2x2 linear system

For fitting a parabola through given points, we’ll solve systems using the method of elimination.
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Example A.4.29.
Solve the system by elimination.
\begin{align*} 5x-2y \amp = 22\\ 2x-5y \amp = 13 \end{align*}
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Solution.
To eliminate the \(x\)-terms,look for the smallest integer that both 2 and 5 divide into evenly, namely, 10. Multiply the first equation by 2 and the second equation by \(-5\text{.}\)
\begin{align*} \blert{2}(5x-2y \amp = 22) \amp\amp \rightarrow \amp 10x-4y \amp = 44\\ \blert{-5}(2x-5y \amp = 13) \amp\amp \rightarrow \amp -10x+25y \amp = -65 \end{align*}
Add these new equations to obtain an equation in \(y\text{.}\)
\begin{align*} 10x-~4y \amp = 44\\ \underline{-10x+25y} \amp \underline{{}= -65 \vphantom{-10x+25y}}\\ 21y \amp = -21 \end{align*}
Solve for \(y\) to find \(y=-1\text{.}\) Finally, substitute \(y=\alert{-1}\) into the first equation and solve for \(x\text{.}\)
\begin{align*} 5x-2(\alert{-1}) \amp = 22\\ 5x+2 \amp = 22\\ x \amp = 4 \end{align*}
The solution to the system is \((4,-1)\text{.}\)
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Checkpoint A.4.30.
Solve the system by elimination.
\begin{align*} 2x-9y \amp = 3\\ 4x-5y \amp = -7 \end{align*}
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Answer.
\((-3,-1)\)
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Checkpoint A.4.31.
Solve the system by elimination.
\begin{align*} 5x+2y \amp = 5\\ 4x+3y \amp = -3 \end{align*}
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Answer.
\((3,-5)\)
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Subsubsection A.4.3.3 Special 3x3 linear system

In this special case of solving a 3x3 system, we can eliminate \(c\) to create a 2x2 system.
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Example A.4.32.
Solve the system by elimination.
\begin{align*} a+b+c \amp = 3 \amp \amp \text{(1)}\\ 4a-2b+c \amp = 18 \amp \amp \text{(2)}\\ 9a+3b+c \amp = 13 \amp \amp \text{(3)} \end{align*}
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Solution.
Eliminate \(c\) by subtracting (1) from (2), then eliminate \(c\) again by subtracting (1) from (3), to get a 2x2 system:
\begin{align*} 3a-3b \amp = 15 \\ 8a+2b \amp = 10 \end{align*}
Divide the first equation by 3 and the second equation by 2, then add.
\begin{align*} a-b \amp = 5 \\ \underline{4a+b} \amp \underline{{}= 5 \vphantom{4a+b}} \\ 5a \amp = 10 \end{align*}
We see that \(a=2\text{.}\) Substituting \(a=2\) into the equation \(a-b=5\text{,}\) we find that \(b=-3\text{.}\) Finally, we substitue \(a=2\) and \(b=-3\) into equation (1) to find
\begin{align*} 2-3+c \amp = 3 \\ c \amp = 4 \end{align*}
The solution is \(a=2,~b=-3\text{,}\) and \(c=4\text{.}\)
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Checkpoint A.4.33.
Solve the system by elimination.
\begin{align*} a+b+c \amp = 5\\ 4a-2b+c \amp = -7\\ 16a+4b+c \amp = -37 \end{align*}
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Answer.
\(a=-3,~b=1,~c=7\)
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Subsection A.4.4 Quadratic Inequalities

Subsubsection A.4.4.1 Solve a linear inequality

First, let’s review solving linear inqualities.
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Example A.4.34.
Solve \(~-3x+1 \gt 7~\) and graph the solutions on a number line.
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Solution.
\begin{align*} -3x+1 \amp \gt 7 \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ -3x \amp \gt 6 \amp \amp \blert{\text{Divide both sides by -3.}}\\ x \amp \lt -2 \amp \amp \blert{\text{Reverse the direction of the inequality.}} \end{align*}
The graph of the solutions is shown below.
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number line
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Example A.4.35.
Solve \(~-3 \lt 2x-5 \le 6~\) and graph the solutions on a number line.
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Solution.
\begin{align*} -3 \amp \lt 2x-5 \le 6 \amp \amp \blert{\text{Add 5 on all three sides.}}\\ 2 \amp \lt 2x \le 11 \amp \amp \blert{\text{Divide each side by 2.}}\\ 1 \amp \lt x \le \dfrac{11}{2} \amp \amp \blert{\text{Do not reverse the inequality.}} \end{align*}
The graph of the solutions is shown below.
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number line
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Checkpoint A.4.36.
Solve the inequality \(~8-4x \gt -2~\)
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Answer.
\(x \lt \dfrac{5}{2}\)
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Checkpoint A.4.37.
Solve the inequality \(~-6 \le \dfrac{4-x}{3} \lt 2~\)
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Answer.
\(22 \ge x \gt -2\)
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Checkpoint A.4.38.
Solve the inequality \(~3x-5 \lt -6x+7~\)
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Answer.
\(x \lt \dfrac{4}{3}\)
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Checkpoint A.4.39.
Solve the inequality \(~-6 \gt 4-5b \gt -21~\)
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Answer.
\(2 \lt b \lt 5\)
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Subsubsection A.4.4.2 \(x\)-intercepts of a parabola

To solve a quadratic inequality, we first find the \(x\)-intercepts of the graph. Remember that there are four different methods for solving a quadratic equation.
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Example A.4.40.
Find the \(x\)-intercepts of the parabola \(~y=4x^2-12\)
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Solution.
Set \(y=0\) and solve for \(x\text{.}\) Use extraction of roots.
\begin{align*} 4x^2-12 \amp = 0 \\ 4x^2 \amp = 12 \\ x^2 \amp = 3 \\ x \amp = \pm \sqrt{3} \end{align*}
The \(x\)-intercepts are \((\sqrt{3},0)\) and \((-\sqrt{3},0)\text{,}\) or about \((1.7,0)\) and \((-1.7,0)\text{.}\)
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Example A.4.41.
Find the \(x\)-intercepts of the parabola \(~y=-4x^2-12x\)
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Solution.
Set \(y=0\) and solve for \(x\text{.}\) Factor the right side.
\begin{align*} 0 \amp = -4x^2-12x \\ 0 \amp = -4x(x+3) \amp \amp \blert{\text{Set each factor equal to 0.}}\\ 4x\amp = 0~~~~~~x+3 = 0 \\ x \amp = 0~~~~~~x=-3 \end{align*}
The \(x\)-intercepts are \((0,0)\) and \((-3,0)\text{.}\)
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Example A.4.42.
Find the \(x\)-intercepts of the parabola \(~y=4x^2-12x+8\)
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Solution.
Set \(y=0\) and solve for \(x\text{.}\) Factor the right side.
\begin{align*} 0 \amp = 4x^2-12x+8 \\ 0 \amp = 4(x^2-3x+2) \\ 0 \amp = 4(x-2)(x-1) \amp \amp \blert{\text{Set each factor equal to 0.}}\\ x-2\amp = 0~~~~~~x-1 = 0 \\ x \amp = 2~~~~~~x=1 \end{align*}
The \(x\)-intercepts are \((2,0)\) and \((1,0)\text{.}\)
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Example A.4.43.
Find the \(x\)-intercepts of the parabola \(~y=12-12x-4x^2\)
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Solution.
Set \(y=0\) and solve for \(x\text{.}\) Use the quadratic formula.
\begin{align*} 0 \amp = -4x^2-12x+12 \amp \amp \blert{a=-4,~b=-12,~c=12}\\ x \amp = \dfrac{12 \pm \sqrt{(-12)^2-4(-4)(12)}}{2(-4)}\\ \amp = \dfrac{12 \pm \sqrt{144+96}}{-8}\\ \amp = \dfrac{12 \pm \sqrt{240}}{-8} = \dfrac{12 \pm 4\sqrt{15}}{-8} \amp \amp \blert{\sqrt{240} = \sqrt{16 \cdot 15} = 4\sqrt{15}}\\ \amp =\dfrac{-3 \pm \sqrt{15}}{2} \end{align*}
The \(x\)-intercepts are \(\left(\dfrac{-3 + \sqrt{15}}{2},0\right)\) and \(\left(\dfrac{-3 - \sqrt{15}}{2},0\right)\text{,}\) or about \((0.44,0)\) and \((-3.44,0)\text{.}\)
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For each Exercise, find the \(x\)-intercepts of the parabola.
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Checkpoint A.4.44.
\(~y=2x^2-7x+3\)
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Answer.
\(\left(\dfrac{1}{2},0\right),~(3,0)\)
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Checkpoint A.4.45.
\(~y=7x-2x^2\)
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Answer.
\((0,0)\text{,}\) \(\left(\dfrac{7}{2},0\right)\)
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Checkpoint A.4.46.
\(~y=10-2x^2\)
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Answer.
\((2.24,0),~(-2.24,0)\)
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Checkpoint A.4.47.
\(~y=2x^2+10x+3\)
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Answer.
\((-0.32,0),~(-4.68,0)\)
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