Question

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

Answer 1

 i think the answer is y=-x-1=-(x+1)

Answer 2

`x^2 = 4y \Rightarrow y= (x^2)/(4)`

`(dy)/(dx) = (2x)/(4) = (x)/(2)`

(x = -2)
`(dy)/(dx) = (x)/(2) = (-2)/(2) = -1`
(x = -2)
`y= (x^2)/(4)= ((-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)`