Find the cubic function y = x^3 + ax^2 + b^x + c whose graph has a horizontal tangent at (-2, 10) and passes through (1,1).
What I tried so far is plugging in -2 and 1 in the original function to get:
-8 +4a -2b +c = 10
1 + a + b + c = 1
And I have y' = 3x^2 + 2ax + b, which when I plug in -2 I get
0 = 12 - 4a + b (so b = -12 + 4a), but I have no clue where to go from there..even though I have a feeling it's something easy.
So if someone could give me a hint as to where to go next, that would be great! thanks for your time
You're off to a good start. I would write the 3 resulting equations as:
(1)
(2)
(3)
Now, if you subtract (2) from (1), you get:
or
Now, if you subtract this from (3) you get:
or and this means
and then substituting into (2) we find
I can't believe I forgot how to find three unknowns from three equations >.>
Well thanks for clearing it up, after that and a quick tutorial from Khan Academy I was quickly able to find the solution!
It turned out to be:
y = x^3 + 2x^2 - 4x + 2
Also thanks to the person above me for working it out, now I'm certain it is correct