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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.
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 \(~~2+x\text{?}\)
  2. Is \(~~\sqrt{\dfrac{x^2}{9}}~~\) equivalent to \(~~\dfrac{x}{3}~~\) for \(~x \ge 0\text{?}\)
  3. Is \(~~\sqrt{w-3}~~\) equivalent to \(~~\sqrt{w} - \sqrt{3}\text{?}\)
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.
  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.
  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.
Decide whether the expressions are equivalent. Assume all variables are positive.
Checkpoint A.4.2.
\(\sqrt{b^2-81}~\) and \(~b-9\)
Answer.
No
Checkpoint A.4.3.
\(\sqrt{64x^2y^2}~\) and \(~8xy\)
Answer.
Yes
Checkpoint A.4.4.
\(\sqrt{64+x^2y^2}~\) and \(~8+xy\)
Answer.
No
Checkpoint A.4.5.
\(\sqrt{\dfrac{c^2+d^2}{4b^2}}~\) and \(~\dfrac{\sqrt{c^2+d^2}}{2b}\)
Answer.
Yes

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.
Example A.4.6.
Simplify each expression. Do not use a calculator!
  1. \(\displaystyle \dfrac{4-\sqrt{64}}{2}\)
  2. \(\displaystyle -2\left(3\sqrt{16}-\sqrt{3(27)}\right)\)
  3. \(\displaystyle 6-3\sqrt[3]{27-7(5)}\)
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*}
  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*}
  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*}
Example A.4.7.
Simplify each expression. Round your answer to hundredths.
  1. \(\displaystyle \dfrac{8-2\sqrt{2}}{4}\)
  2. \(\displaystyle 2+6\sqrt[3]{-25}\)
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.
    On a calculator, enter
    \(\qquad\qquad\) ( \(8\) - \(2~\boxed{\sqrt{}}\) ) ) ÷ \(4\) ENTER
    and round to two decimal places: \(~1.29\)
  2. Evaluate the cube root, multiply by 6, then add the result to 2.
    On a calculator, enter
    \(\qquad\qquad 2\) + \(6\) MATH 4 - \(25\) ENTER
    and round to two decimal places: \(~{-15.54}\)
Checkpoint A.4.8.
Simplify each expression. Do not use a calculator!
  1. \(\displaystyle \dfrac{36}{6+\sqrt{36}}\)
  2. \(\displaystyle 10+2(3-\sqrt{169})\)
  3. \(\displaystyle \dfrac{3+\sqrt[3]{-729}}{6-\sqrt[3]{-27}}\)
Answer.
  1. \(\displaystyle 3\)
  2. \(\displaystyle -10\)
  3. \(\displaystyle \dfrac{-2}{3}\)
Checkpoint A.4.9.
Simplify each expression. Round your answer to hundredths.
  1. \(\displaystyle \dfrac{6+9\sqrt{3}}{3}\)
  2. \(\displaystyle -1-3\sqrt[3]{120}\)
Answer.
  1. \(\displaystyle 7.20\)
  2. \(\displaystyle -15.80\)

Subsubsection A.4.1.3 Use the quadratic formula

Example A.4.10.
  1. Use the quadratic formula to solve the equation \(~x = \dfrac{2}{3} - \dfrac{x^2}{6}\)
  2. Give decimal approximations to two decimal places for the solutions.
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 \(\dfrac{-6 - \sqrt{52}}{2}\)
  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 \(-6.61\text{.}\)
Checkpoint A.4.11.
Use the quadratic formula to solve the equation \(x^2 + 2x = 5\)
Answer.
\(x = \dfrac{-2 \pm \sqrt{24}}{2}\)

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.
Example A.4.12.
Graph the equation \(~y=2(x-15)^2-72\)
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.
parabola
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.
Answer.
parabola

Subsubsection A.4.2.2 Product of binomials

With practice, you will be able to multiply binomials mentally.
Example A.4.14.
Multiply \(~(x-6)(x+3)\text{.}\)
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*}
Example A.4.15.
Expand \(~(2t-5)^2\text{.}\)
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*}
Checkpoint A.4.16.
Multiply \(~(4a-7)(3a+8)\text{.}\)
Answer.
\(12a^2+11a-56\)
Checkpoint A.4.17.
Expand \(~(3-4b)^2\text{.}\)
Answer.
\(16b^2-24b+9\)

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.
Example A.4.18.
The point \((6,2)\) lies on the graph of \(~y=a(x-4)^2+1.~\) Solve for \(a\text{.}\)
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 \(a = \dfrac{1}{4}\text{.}\)
Example A.4.19.
The point \((-2,11)\) lies on the graph of \(~y=x^2+bx-3.~\) Solve for \(b\text{.}\)
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 \(b = -5\text{.}\)
Checkpoint A.4.20.
The point \((-6,10)\) lies on the graph of \(~y=a(x+3)^2-2.~\) Solve for \(a\text{.}\)
Answer.
\(a=\dfrac{4}{3}\)
Checkpoint A.4.21.
The point \((-3,8)\) lies on the graph of \(~y=-x^2+bx+5.~\) Solve for \(b\text{.}\)
Answer.
\(b=-4\)
Checkpoint A.4.22.
The point \((8,-12)\) lies on the graph of \(~y=ax^2-4x+36.~\) Solve for \(a\text{.}\)
Answer.
\(a=\dfrac{-1}{3}\)
Checkpoint A.4.23.
The point \((60,-480)\) lies on the graph of \(~y=\dfrac{-2}{3}(x-h)^2+120.~\) Solve for \(h\text{.}\)
Answer.
\(h=30\)

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.
Example A.4.24.
Write an equation to say that \((-3,8)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
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*}
Checkpoint A.4.25.
Write an equation to say that \((-4,-18)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
Answer.
\(-16a-4b+c=-18\)
Checkpoint A.4.26.
Write an equation to say that \((8,0)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
Answer.
\(64a+8b+c=0\)
Checkpoint A.4.27.
Write an equation to say that \((0,-5)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
Answer.
\(c=-5\)
Checkpoint A.4.28.
Write an equation to say that \((-60,400)\) lies on the graph of\(~y=ax^2+bx+c~\text{.}\)
Answer.
\(3600a-60b+c=400\)

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.
Example A.4.29.
Solve the system by elimination.
\begin{align*} 5x-2y \amp = 22\\ 2x-5y \amp = 13 \end{align*}
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 \((4,-1)\text{.}\)
Checkpoint A.4.30.
Solve the system by elimination.
\begin{align*} 2x-9y \amp = 3\\ 4x-5y \amp = -7 \end{align*}
Answer.
\((-3,-1)\)
Checkpoint A.4.31.
Solve the system by elimination.
\begin{align*} 5x+2y \amp = 5\\ 4x+3y \amp = -3 \end{align*}
Answer.
\((3,-5)\)

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.
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*}
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 \(c=4\text{.}\)
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*}
Answer.
\(a=-3,~b=1,~c=7\)

Subsection A.4.4 Quadratic Inequalities

Subsubsection A.4.4.1 Solve a linear inequality

First, let’s review solving linear inqualities.
Example A.4.34.
Solve \(~-3x+1 \gt 7~\) and graph the solutions on a number line.
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.
number line
Example A.4.35.
Solve \(~-3 \lt 2x-5 \le 6~\) and graph the solutions on a number line.
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.
number line
Checkpoint A.4.36.
Solve the inequality \(~8-4x \gt -2~\)
Answer.
\(x \lt \dfrac{5}{2}\)
Checkpoint A.4.37.
Solve the inequality \(~-6 \le \dfrac{4-x}{3} \lt 2~\)
Answer.
\(22 \ge x \gt -2\)
Checkpoint A.4.38.
Solve the inequality \(~3x-5 \lt -6x+7~\)
Answer.
\(x \lt \dfrac{4}{3}\)
Checkpoint A.4.39.
Solve the inequality \(~-6 \gt 4-5b \gt -21~\)
Answer.
\(2 \lt b \lt 5\)

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.
Example A.4.40.
Find the \(x\)-intercepts of the parabola \(~y=4x^2-12\)
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 \((-1.7,0)\text{.}\)
Example A.4.41.
Find the \(x\)-intercepts of the parabola \(~y=-4x^2-12x\)
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 \((-3,0)\text{.}\)
Example A.4.42.
Find the \(x\)-intercepts of the parabola \(~y=4x^2-12x+8\)
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 \((1,0)\text{.}\)
Example A.4.43.
Find the \(x\)-intercepts of the parabola \(~y=12-12x-4x^2\)
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 \((-3.44,0)\text{.}\)
For each Exercise, find the \(x\)-intercepts of the parabola.
Checkpoint A.4.44.
\(~y=2x^2-7x+3\)
Answer.
\(\left(\dfrac{1}{2},0\right),~(3,0)\)
Checkpoint A.4.45.
\(~y=7x-2x^2\)
Answer.
\((0,0)\text{,}\) \(\left(\dfrac{7}{2},0\right)\)
Checkpoint A.4.46.
\(~y=10-2x^2\)
Answer.
\((2.24,0),~(-2.24,0)\)
Checkpoint A.4.47.
\(~y=2x^2+10x+3\)
Answer.
\((-0.32,0),~(-4.68,0)\)
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