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MATH 120 -- EXAM I Hints and Answers

Version 1

Autumn 2005
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\begin{enumerate}

\item ANSWER:  $4x+2h$

\item HINT:  The quadratic equation $ax^2+bx+c=0$ will have exactly one solution if $b^2-4ac=0$.  

ANSWER:  $\alpha=\pm\sqrt{3}$

\item HINT: With a coordinate system imposed so that the origin is at Pam's apartment building, the circle that defines the range of the ``hotspot" is $(x-3)^2+(y-0.25)^2=0.3125^2.$  (The radius of the circle is 1650~feet, which is 0.3125~miles.)  The circle intersects the $x$-axis at $x=3\pm0.1875$.  So, Pam is in the circle for a distance of $2\times 0.1875=0.375$~miles.  Use the fact that the bus travels 40~mph to find the amount of time she is in the circle in hours and convert to minutes.

ANSWER:  0.5625 minutes

\item (5 points each)

\begin{enumerate}

\item HINT:  Enrollment reaches a maximum of 3600 students at time $t=10$.  This means that the vertex of the parabola is at the point $(10,3600)$.  That gives the values of $h$ and $k$.  So, $C(t)=a(t-10)^2+3600$.  To find $a$, use the fact that $C(0)=a(0-10)^2+3600$ and $C(0)=852$.  That gives the equation $100a+3600=852.$  Solve for $a$.

ANSWER:  $h=10$, $k=3600$, $a=-27.48$

\item HINT:  Find the equation of the line through the points $(0,200)$ and $(10,950)$.

ANSWER:  $L(t)=75t+200$

\item HINT:  The right-handed population is the difference between the total enrollment and the left-handed population.  That is, $R(t)=C(t)-L(t)$, where $C(t)=-27.48(t-10)^2+3600$ and $L(t)=75t+200$.  After some simplification, $R(t)=-27.48t^2+474.6t+652$.  This is a quadratic, whose graph is a parabola that opens down.  It is largest at its vertex.

ANSWER:  $t=8.64$ years

\end{enumerate}

\item (5 points each)

\begin{enumerate}

\item HINT:  The $y$-intercept of the graph is $f(0)=(0)^2+6(0)+8=0$.  So, the line segment goes through the points $(0,8)$ and $(9,0)$.  Find its equation.

ANSWER:  $-\frac89x+8$

\item HINT:  For the range, we need to know the lowest and highest $y$-values on the graph of $f$.  The high value occurs at either $x=-6$ or $x=0$.  It turns out that $f(-6)$ and $f(0)$ are both 8.  The low value occurs at either $x=10$ or the vertex of the parabola.  The vertex of the parabola occurs at $x=-3$.  (Why?)  Further, $f(-3)=(-3)^2+6(-3)+8=-1$ and $f(10)=-\frac89(10)+8=-\frac89$.  So, the low point on the graph has $y$-value $-1$.

ANSWER:  $D=\{x|-6\leq x \leq 10\}$; $R=\{y|-1\leq y \leq 8\}$

\item HINT:  $(g\circ f)(x)=\sqrt{f(x)-3}$.  The domain of this function consists of all values of $x$ that are in the domain of $f$ and such that $f(x)-3\geq 0$.  That is, the domain of $g\circ f$ consists of all values of $x$ such that $-6\leq x \leq 10$ and $f(x)\geq3$.  From the graph of $f$, I can see that $f(x)=3$ in three places.  Setting $x^2+6x+8=3$ gives me two of these places:  $x=-5$ and $x=-1$;  setting $-\frac89x+8=3$ gives me the other.  From the graph, I can see that $f(x)\geq 3$ if $-6\leq x\leq -5$ or $-1\leq x\leq \frac{45}8$.

ANSWER:  $D_{g\circ f}=[-6,-5]\cup[-1,\frac{45}8]$

\end{enumerate}

\end{enumerate}

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