Answer:
[tex]\frac{3}{16} x^{3}-\frac{9}{4}x+6 = y[/tex]
Step-by-step explanation:
Lets start with the easiest part of this question - simply plugging in to get easier equations:
[tex]-8a+4b-2c+d=9\\[/tex]
[tex]8a+4b+2c+d=3[/tex]
Something that now immedietly pops out is that many of the terms would cancel if we added them together. We get the following results by adding the two equations together:
[tex]8b+2d = 12[/tex]
Now lets look at the other bit of inromation the question gives you -- that the two points are horizontal tangents. Taking the derivative of the standard form of a cubic, we get:
[tex]y' = 3ax^2+2bx+c[/tex]
Since the points are both horizontal tangents, y' will be equal to 0 at the points. Thus, plugging in we get:
[tex]y'(-2) = 0 = 12a-4b+c[/tex]
[tex]y'(2) = 0 = 12a+4b+c[/tex]
We again see a simple subtracting opportunity that will cancel out two terms:
[tex]-8b = 0[/tex]
[tex]b = 0[/tex]
Now going back to the equation we got by adding the two equations that we got from simply plugging our points in:
[tex]d = 6[/tex]
Now that we have b and d, we can now plug them back into our step one and derivative equations in order to get a simple system:
[tex]-8a - 2c=3[/tex]
[tex]12a+c=0[/tex]
Solving for a and c, we get:
[tex]a = \frac{3}{16}[/tex]
[tex]c = \frac{-9}{4}[/tex]
Thus, finally, our answer is:
[tex]\frac{3}{16} x^{3}-\frac{9}{4}x+6 = y[/tex]
Attached is desmos, which you can use to check.
