```Question 113917
Since the zeros are {{{x=-3}}} and {{{x=5}}} we can find the factorization using these zeros. We simply need to use the zero product property in reverse

{{{x+3=0}}} or {{{x-5=0}}} Get the numbers to the left side

Since either piece equals zero, then their product equals zero

{{{(x+3)(x-5)=0}}}

However, this equation may not pass through (3,6). So let's introduce an "a" coefficient in front to get the equation:

{{{y=a(x+3)(x-5)}}}

Now since the equation passes through (3,6), this means x=3 and y=6

{{{6=a(3+3)(3-5)}}} Plug in x=3 and y=6

{{{6=a(6)(-2)}}} Combine like terms

{{{6=a(-12)}}} Multiply

{{{6/(-12)=a}}} Divide both sides by -12

{{{-1/2=a}}} Reduce

So our answer is {{{a=-1/2}}} which means the equation is {{{y=(-1/2)(x+3)(x-5)}}}

Notice if we graph {{{y=(-1/2)(x+3)(x-5)}}} and plot the point (3,6), we can see that the point lies on the line and {{{y=(-1/2)(x+3)(x-5)}}} has zeros -3 and 5. So this verifies our answer.

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(-1/2)(x+3)(x-5))
circle(3,6,0.08),
circle(3,6,0.1),
circle(3,6,0.15)
)}}}```