Question 236025
Since f(0)=-1, f(3)=-10 this means we have the two points (0,-1) and (3,-10)




First let's find the slope of the line through the points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(3,-10\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(0,-1\right)]. So this means that {{{x[1]=0}}} and {{{y[1]=-1}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,-10\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=-10}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(-10--1)/(3-0)}}} Plug in {{{y[2]=-10}}}, {{{y[1]=-1}}}, {{{x[2]=3}}}, and {{{x[1]=0}}}



{{{m=(-9)/(3-0)}}} Subtract {{{-1}}} from {{{-10}}} to get {{{-9}}}



{{{m=(-9)/(3)}}} Subtract {{{0}}} from {{{3}}} to get {{{3}}}



{{{m=-3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(3,-10\right)] is {{{m=-3}}}



Now let's use the point slope formula:



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y--1=-3(x-0)}}} Plug in {{{m=-3}}}, {{{x[1]=0}}}, and {{{y[1]=-1}}}



{{{y+1=-3(x-0)}}} Rewrite {{{y--1}}} as {{{y+1}}}



{{{y+1=-3x+-3(-0)}}} Distribute



{{{y+1=-3x+0}}} Multiply



{{{y=-3x+0-1}}} Subtract 1 from both sides. 



{{{y=-3x-1}}} Combine like terms. 



{{{y=-3x-1}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(3,-10\right)] is {{{y=-3x-1}}}



 Notice how the graph of {{{y=-3x-1}}} goes through the points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(3,-10\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-3x-1),
 circle(0,-1,0.08),
 circle(0,-1,0.10),
 circle(0,-1,0.12),
 circle(3,-10,0.08),
 circle(3,-10,0.10),
 circle(3,-10,0.12)
 )}}} Graph of {{{y=-3x-1}}} through the points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(3,-10\right)]