What is the equation of the normal line to the curve y = x^2 + 2x -1 at point (1,2)
attempt: dy/dx = 2x + 0
dy/dx = 2x,
im stuck here
Your derivative is not quite right:
$\displaystyle y=x^2+2x-1$
$\displaystyle \frac{dy}{dx}=2x+2$
The slope m of the normal line, however, is:
$\displaystyle m=-\frac{dx}{dy}=-\frac{1}{\frac{dy}{dx}}$
So, for $\displaystyle x=1$, what is the slope of the normal line?
You found the slope of the tangent line $\displaystyle m_{\text{T}}$, but you need the slope of the normal line $\displaystyle m_{\text{N}}$. As Prove It pointed out, you require:
$\displaystyle m_{\text{N}}m_{\text{T}}=-1$
A normal line is perpendicular to the curve at the point of intersection, thus it is perpendicular to the tangent line at that point.
As far as the product of perpendicular slopes being -1, consider that the difference in their angles of inclination must be $\displaystyle \frac{\pi}{2}$:
$\displaystyle \tan^{-1}(m_1)-\tan^{-1}(m_2)=\frac{\pi}{2}$
Using the identity $\displaystyle \tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}$ we may write:
$\displaystyle \tan^{-1}(m_1)-\tan^{-1}(m_2)=\tan^{-1}(m_1)+\cot^{-1}(m_1)$
$\displaystyle \tan^{-1}(-m_2)=\cot^{-1}(m_1)$
$\displaystyle \tan^{-1}(-m_2)=\tan^{-1}\left(\frac{1}{m_1} \right)$
$\displaystyle m_1m_2=-1$