Q. 9 - Precalculus Enhanced with Graphing Utilities

Question

Use the given graph of the function $f$ to answer parts (a) – (n).

(a) Find $f\left(0\right)$ and $f\left(-6\right)$.

(b) Find $f\left(6\right)$ and $f\left(11\right)$.

(c) Is $f\left(3\right)$ positive or negative?

(d) Is $f\left(-4\right)$ positive or negative?

(e) For what values of $x$ is $f\left(x\right)=0$?

(f) For what values of $x$ is $f\left(x\right)>0$?

(g) What is the domain of $f$?

(h) What is the range of $f$?

(i) What are the x-intercepts?

(j) What is the y-intercept?

(k) How often does the line $y=\frac{1}{2}$ intersect the graph?

(l) How often does the line $x=5$ intersect the graph?

(m) For what values of $x$ does $f\left(x\right)=3$?

(n) For what values of $x$ does $f\left(x\right)=-2$? 

Step-by-Step Solution

Answer

(a) $f\left(0\right)=3; f\left(-6\right)=-3$

(b) $f\left(6\right)=0; f\left(11\right)=1$

(c) Positive

(d) Negative

(e) The values of $x$ are $-3, 6, and 10$.

(f) The values of $x$ are  $-3<x<6 or 10<x\le 11$.

(g)  $\left\{x\overline{)-6\le x\le 11}\right\}$

(h)  $\left\{y\overline{)-3\le y\le 4}\right\}$

(i) The x-intercepts are $-3, 6, and 10$.

(j) The y-intercept is $3$.

(k) $3$ times

(l) Once

(m) The values of $x$ are $0 and 4$.

(n) The values of $x$ are $-5 and 8$.

Part (a) Step 1.

Since $\left(0, 3\right)$ is on the graph of $f$, the y-coordinate $3$ is the value of $f$ at the x-coordinate $0$; that is, $f\left(0\right)=3$. When $x=-6$, then $y=-3$, so $f\left(-6\right)=-3$. 

Part (b) Step 1.

Since $\left(6, 0\right)$ is on the graph of $f$, the y-coordinate $0$ is the value of $f$ at the x-coordinate $6$; that is, $f\left(6\right)=0$. When $x=11$, then $y=1$, so $f\left(11\right)=1$. 

Part (c) Step 1.

The value of $f\left(3\right)$ is between $3$ and $4$. So, $f\left(3\right)$ is positive.

Part (d) Step 1.

The value of $f\left(-4\right)$ is between $-2$ and $0$. So, $f\left(-4\right)$ is negative.

Part (e) Step 1.

Since $\left(-3, 0\right)$, $\left(6, 0\right)$ and $\left(10, 0\right)$ are the only points on the graph for which $y=f\left(x\right)=0$. So, $f\left(x\right)=0$   when $x=-3$, $x=6$ and $x=10$. 

Part (f) Step 1.

First, determine the x-values from $-6$ to $11$ for which the y-coordinate is positive. $f\left(x\right)>0$ occurs for  $-3<x<6 or 10<x\le 11$.

Part (g) Step 1.

The points on the graph of $f$ have x-coordinates between $-6$ and $11$, inclusive; and for each number x between $-6$ and $11$, there is a point $\left(x, f\left(x\right)\right)$ on the graph. The domain of $f$ is $\left\{x\overline{)-6\le x\le 11}\right\}$.

Part (h) Step 1.

The points on the graph all have y-coordinates between $-3$ and $4$, inclusive; and for each such number $y$, there is at least one corresponding number $x$ in the domain. The range of $f$ is $\left\{y\overline{)-3\le y\le 4}\right\}$.

Part (i) Step 1.

The x-intercepts of the graph of $f$ are the real solutions of the equation $f\left(x\right)=0$ that are in the domain of $f$.

So, the x-intercepts are $-3, 6 and 10$.

Part (j) Step 1.

The y-intercepts of the graph of $f$ are the real values of $f\left(0\right)$.

So, the y-intercept is $3$.

Part (k) Step 1.

If a horizontal line $y=\frac{1}{2}$ is drawn on the graph then it intersects the graph three times.

Part (l) Step 1.

If a vertical line $x=5$ is drawn on the graph then it intersects the graph only once.

Part (m) Step 1.

Since $\left(0, 3\right)$ and $\left(4, 3\right)$ are the only points on the graph for which $y=f\left(x\right)=3$. So, $f\left(x\right)=3$ when $x=0$ and $x=4$.

Part (n) Step 1.

Since $\left(-5,-2\right)$ and $\left(8,-2\right)$ are the only points on the graph for which $y=f\left(x\right)=-2$. So, $f\left(x\right)=-2$ when $x=-5$ and $x=8$.