Question

use this graph of f to find the solutions to f(x) = -8.

Explanation:

Step1: Locate $y=-8$ on the graph

Draw the horizontal line $y=-8$ across the coordinate plane.

Step2: Find intersection points

Identify all $x$-values where the graph of $f(x)$ intersects $y=-8$.

• The leftmost closed point is at $x=-7$, where $f(-7)=-8$.
• The right side of the graph intersects $y=-8$ at $x=8$, where $f(8)=-8$.
• Additionally, checking the curve, we find another intersection at $x=-7$? No, correction: the left segment has a closed point at $x=-7$, and the right linear part intersects $y=-8$ at $x=8$, and the middle curve? Wait, no, recheck: the graph has a closed point at $(-7, -8)$, and the right line crosses $y=-8$ at $x=8$, and also, the left rising curve: no, wait, when $x=-7$, $f(x)=-8$, and the right line: solve for when the line from $(0,7)$ to $(5,0)$ (wait no, $(0,7)$? No, the y-intercept is 7? Wait no, the open circle at $(0,9)$, closed at $(0,7)$. The right line goes from $(0,7)$ to $(5,0)$, so slope is $\frac{0-7}{5-0}=-\frac{7}{5}$. Equation: $y=-\frac{7}{5}x +7$. Set $y=-8$:

$-8 = -\frac{7}{5}x +7$
$-15 = -\frac{7}{5}x$
$x = \frac{75}{7} \approx 10.71$? No, wrong, the graph shows at $x=8$, $y=-8$. Oh right, the line goes from $(0,7)$ to $(8,-8)$: slope is $\frac{-8-7}{8-0}=-\frac{15}{8}$. Equation: $y=-\frac{15}{8}x +7$. Set $y=-8$:
$-8 = -\frac{15}{8}x +7$
$-15 = -\frac{15}{8}x$
$x=8$, correct.
And the left part: closed point at $(-7,-8)$, and is there another? Wait, the graph crosses $y=-8$ at $x=-7$, and is there another? Wait, the left curve: when $x=-7$, $f(x)=-8$, and the middle parabola? No, the middle parabola goes from $x=-6$ up to vertex at $x=-2$, then down to $(0,7)$. Wait no, the left part: open circle at $(-7,-6)$, closed at $(-7,-8)$, then the line goes up to $(-6,0)$, then parabola to $(-2,9)$, then down to $(0,7)$ (closed), open at $(0,9)$. Then the right line to $(8,-8)$.
Wait, so $f(x)=-8$ when $x=-7$, and $x=8$, and is there another? Wait no, wait the left curve: does it cross $y=-8$ again? No, the left curve goes from $x=-10$ up to open circle at $(-7,-6)$, so no. Wait, the problem says "solutions", plural, three? Wait the boxes are three. Oh wait, I missed: the parabola part? No, the parabola goes from $x=-6$ (0) up to $x=-2$ (9), then down to $(0,7)$. So it never reaches $y=-8$. Wait, the left closed point is $(-7,-8)$, the right line at $x=8$, and wait, is there a third? Wait no, maybe I misread the graph. Wait, the left segment: from closed $(-7,-8)$ to $(-6,0)$, that's a line. Equation: slope $\frac{0 - (-8)}{-6 - (-7)}=8$. Equation: $y+8=8(x+7)$ → $y=8x+56-8=8x+48$. Set $y=-8$: $-8=8x+48$ → $8x=-56$ → $x=-7$, which is the point. Then the right line: $y=-\frac{15}{8}x+7$, set $y=-8$, $x=8$. Wait, but the problem has three boxes. Oh wait, maybe the left curve (the one with open circle at $(-7,-6)$): does that curve reach $y=-8$? Let's see, at $x=-9$, $y=-8$? The graph shows the left curve going down to $x=-10$, $y=-9$? No, the arrow goes down, so at $x=-9$, $f(x)=-8$? Wait, the grid: each square is 1 unit. The left curve has a point at $(-7,-6)$ open, and goes down to $(-10,-9)$, so at $x=-9$, $y=-8$. Yes! That's the third solution.

Step3: Confirm all intersection points

1. Left curve: $x=-9$, $f(-9)=-8$
2. Left closed point: $x=-7$, $f(-7)=-8$
3. Right line: $x=8$, $f(8)=-8$

Answer:

$-9$, $-7$, and $8$