Find the cubic function y = x3 + ax2 + bx + c

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 :)

Re: Find the cubic function y = x3 + ax2 + bx + c

Quote:

Originally Posted by

**aimforthehead** 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),

Well, you have three equations in three unknowns.

Re: Find the cubic function y = x3 + ax2 + bx + c

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

Re: Find the cubic function y = x3 + ax2 + bx + c

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 :)