for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?
So we are going to need to solve some equations....
I am going to call $\displaystyle f(x)=y=x^3 $
Starting with $\displaystyle y=x^2+ax+b$ we get
$\displaystyle 1=1^2+a(1)+b \implies a=-b$
$\displaystyle f'(x)=3x^2 \implies f'(1)=3$
Now $\displaystyle y'=2x+a$ using the slope above and the x-coordinate we can solve for a
$\displaystyle 3=2(1)+a \implies a =1$ so $\displaystyle b=-1$
$\displaystyle y=x^2+x-1$
First you will want to find the derivative of: $\displaystyle f(x) = x^3$. Using the power rule, $\displaystyle f'(x) = 3x^2$ Now plug in the 1 to find the slope of the curve at x = 1. $\displaystyle 3(1)^2 = 3$. Now you will want to find the derivative of the equation $\displaystyle g(x) = x^2 + ax + b$ and set that equal to the derivative of $\displaystyle f(x)$. So $\displaystyle g'(x) = 2x + a$ and $\displaystyle f'(x) = 3x^2.$
$\displaystyle 3x^2 = 2x + a$
Substituting in the 1:
$\displaystyle
\\3 = 2 + a
$
$\displaystyle
\\a = 1
$
Now that you have the a-value, you will want to solve for b. Set $\displaystyle f(1) = g(1)$ to get the b.
$\displaystyle
\\ x^3 = x^2 + x + b
$
$\displaystyle
\\ 1 = 1 + 1 + b
$
$\displaystyle
\\ b = -2
$
Finally, the equation of the parabola tangent to the cubic is:
$\displaystyle
g(x) = x^2 + x - 1
$