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Find the equation whose graph is shown below. Write your answer in standard form.

(Standard form is , where  is positive, and , , and  are integers with greatest common divisor 1.)

$$[asy] size(150); real ticklen=3; real tickspace=2; real ticklength=0.1cm; real axisarrowsize=0.14cm; pen axispen=black+1.3bp; real vectorarrowsize=0.2cm; real tickdown=-0.5; real tickdownlength=-0.15inch; real tickdownbase=0.3; real wholetickdown=tickdown; void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool useticks=false, bool complexplane=false, bool usegrid=true) { import graph; real i; if(complexplane) { label("\textnormal{Re}",(xright,0),SE); label("\textnormal{Im}",(0,ytop),NW); } else { label("x",(xright+0.4,-0.5)); label("y",(-0.5,ytop+0.2)); } ylimits(ybottom,ytop); xlimits( xleft, xright); real[] TicksArrx,TicksArry; for(i=xleft+xstep; i 0.1) { TicksArrx.push(i); } } for(i=ybottom+ystep; i 0.1) { TicksArry.push(i); } } if(usegrid) { xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray(0.22),extend=true),p=invisible);//,above=true); yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.22),extend=true), p=invisible);//,Arrows); } if(useticks) { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks("%",TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks("%",TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); } else { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize)); } }; rr_cartesian_axes(-6,6,-6,6); draw((-4.5,6) -- (1.5,-6), red,Arrows(size=0.2cm)); [/asy]$$

Jun 3, 2021
edited by Guest  Jun 3, 2021

#2
+181
+1

Standard form is $y = mx+b$, where m is the slope and b is a constant.

Slope is given by $\frac{y_1 - y_2}{x_1 - x_2}$.

Thus, the slope of these points is $-\frac{2}{3}$.

Plugging into our equation gives $y = -\frac{2}{3}x + b$. Plugging in a point (in this case I used $(-3,3)$), gives $3 = \frac{2}{3} \cdot -3 + b$.

Thus, $b = 5$. As a result, our final equation is $y = -\frac{2}{3}x + 5$

Jun 3, 2021

#1
+1

Here is a diagram:

Jun 3, 2021
#2
+181
+1

Standard form is $y = mx+b$, where m is the slope and b is a constant.

Slope is given by $\frac{y_1 - y_2}{x_1 - x_2}$.

Thus, the slope of these points is $-\frac{2}{3}$.

Plugging into our equation gives $y = -\frac{2}{3}x + b$. Plugging in a point (in this case I used $(-3,3)$), gives $3 = \frac{2}{3} \cdot -3 + b$.

Thus, $b = 5$. As a result, our final equation is $y = -\frac{2}{3}x + 5$

xCorrosive  Jun 3, 2021