# 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

#### Plato

MHF Helper
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*}

#### MarkFL

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 $$\displaystyle 2-4+c=0\,\therefore\,c=2$$