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

Sep 2012
5
0
California
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 :)
 

Plato

MHF Helper
Aug 2006
22,508
8,664
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.
\(\displaystyle \begin{align*}4a-2b+c&=18\\a+b+c &=0\\4a-b &=12 \end{align*}\)
 
Dec 2011
2,314
916
St. Augustine, FL.
You're off to a good start. I would write the 3 resulting equations as:

(1) \(\displaystyle 4a-2b+c=18\)

(2) \(\displaystyle a+b+c=0\)

(3) \(\displaystyle 4a-b=12\)

Now, if you subtract (2) from (1), you get:

\(\displaystyle 3a-3b=18\) or \(\displaystyle a-b=6\)

Now, if you subtract this from (3) you get:

\(\displaystyle 3a=6\) or \(\displaystyle a=2\) and this means \(\displaystyle 2-b=6\,\therefore\,b=-4\)

and then substituting into (2) we find \(\displaystyle 2-4+c=0\,\therefore\,c=2\)
 
Sep 2012
5
0
California
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 :)