Asked by Khushal Sharma

The graph of $y=f(x)$ is shown in red below. Find $f(-2)$.

[asy]
size(150);

real ticklength=0.1cm;
real axisarrowsize=0.14cm;
pen axispen=black+1.3bp;
real vectorarrowsize=0.2cm;
real tickdown=-0.5;
real tickdownlength=-0.15inch;
real tickdownbase=0.3;
real wholetickdown=tickdown;
void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool useticks=false, bool complexplane=false, bool usegrid=true) {
import graph;
real i;
if(complexplane) {
label("$\textnormal{Re}$",(xright,0),SE);
label("$\textnormal{Im}$",(0,ytop),NW);
} else {
label("$x$",(xright+0.4,-0.5));
label("$y$",(-0.5,ytop+0.2));
}

ylimits(ybottom,ytop);
xlimits( xleft, xright);
real[] TicksArrx,TicksArry;

for(i=xleft+xstep; i<xright; i+=xstep) {
if(abs(i) >0.1) {
TicksArrx.push(i);
}
}
for(i=ybottom+ystep; i<ytop; i+=ystep) {
if(abs(i) >0.1) {
TicksArry.push(i);
}
}

if(usegrid) {
xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray(0.72),extend=true),p=invisible);//,above=true);
yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.72),extend=true), p=invisible);//,Arrows);
}
if(useticks) {
xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks("%",TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));
yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks("%",TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

} else {
xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize));
yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize));
}
};

rr_cartesian_axes(-5,5,-3,7);
draw((-4,4)--(-1,-1)--(0,3)--(4,1),red+1.125bp);
dot((-4,4)); dot((4,1));
[/asy]
(Assume grid lines are spaced $1$ unit apart and that the graph is made from line segments.)

Answers

Answered by oobleck
sure, if you say so.
Answered by hooooo
ummm kinda hard to um you know solve
ALSO STOP CHEATING FOR AOPS
Answer
dat is crazy!
Answered by hoooooo
whats crazy
Answered by bing bong
hi
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