Find the equation of the tangent line and the normal line for
y=(1+2x)^2 on point (1,9)
I tried to find the slope by finding f' but i don't think I am doing it right.
The derivative is the slope of the tangent line. The derivative of y = (1+2x)^2 is:
$\displaystyle \frac{dy}{dx} = 4(1+2x)$
Given the point (1, 9) and a slope, you can find the equation of the line by
$\displaystyle y - y_1 = m(x-x_1)$
The equation of the normal line passes thru point (1, 9) but has a slope given by:
$\displaystyle m_{normal} \cdot m_{tangent} = -1$