Question

The equation of the tangent to the curve x^2 = 4y at the point on the curve where x=-2 is?

Answer 1

$x^2 = 4y \Rightarrow y= \frac{x^2}{4}$
$\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$

(x = -2) $\frac{dy}{dx} = \frac{x}{2} = \frac{-2}{2} = -1$
(x = -2) $y= \frac{x^2}{4}= \frac{(-2)^2}{4} = 1$

$y-y_1=m(x-x_1)$
$y-1 = -1(x--2)$
$y-1 = -1(x+2)$
$y-1 = -x-2$
$y=-x-1 = -(x+1)$

Answer 2

 i think the answer is y=-x-1=-(x+1)