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 $f\left(x\right)=-2$?

Short 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 role="math" localid="1647277179901" $-3,6,and10$.

(f) The values of$x$arerole="math" localid="1647278873092" $-3<x<6or10<x\le 11$.

(g)role="math" localid="1647279101990" $\left\{x\overline{)-6\le x\le 11}\right\}$

(h)role="math" localid="1647279302824" $\left\{y\overline{)-3\le y\le 4}\right\}$

(i) The x-intercepts are $-3,6,and10$.

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

(k) $3$times

(l) Once

(m) The values of $x$are $0and4$.

(n) The values of $x$are $-5and8$.

Step-by-step solution

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 forrole="math" localid="1647278891188" $-3<x<6or10<x\le 11$.

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 role="math" localid="1647279600096" $-3,6and10$.

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 role="math" localid="1647280898661" $y=f\left(x\right)=3$. 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 $x=8$.