I dont know if this goes in here but this is a question from my pre-calc class that i could not solve
Find a polynomial whose turning points are at -1, (3 + √7)/4 , and (3 - √7)/4
Any help will be appreciated.
Turning Points occur when the derivative is 0.
Using this information, you can find the derivative.
$\displaystyle \frac{dy}{dx} = (x + 1)\left[x - \left(\frac{3 + \sqrt{7}}{4}\right)\right]\left[x - \left(\frac{3 - \sqrt{7}}{4}\right)\right]$
(set it equal to 0, you should see that the turning points are correct).
Expand, and then you can integrate to find the polynomial.
$\displaystyle \frac{dy}{dx} = x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}$
$\displaystyle y = \int{x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}\,dx}$
$\displaystyle y = \frac{1}{4}x^4 - \frac{1}{6}x^3 - \frac{1}{2}x^2 + \frac{1}{2}x + C$.
You can choose any value of C you like... 0 is the easiest.
So a polynomial that has the turning points you mentioned is
$\displaystyle y = \frac{1}{4}x^4 - \frac{1}{6}x^3 - \frac{1}{2}x^2 + \frac{1}{2}x$.
Why are you doing questions that require integration if you never learned it?
Your answer is wrong. Read this: Integrals of Polynomials