Question

How do you find the equation of a line tangent to `y=-x^2` at x=0?

Answer 1

`y=0`

Explanation:

`y = -x^2`

Differentiating wrt `x` gives:
`dy/dx = -2x`

So when `x=0 => dy/dx = 0`, which is the gradient of the tangent at `x=0`.

Also, when `x=0=>y=0`

So, the equation of the tangent is give by `y-y_1=m(x-x_1)`
`:. y-0=0(x-0) => y=0`

Hence, the equation is `y=0`.

This should come as no surprise, if you think about the graph:
graph{-x^2 [-10, 10, -5, 5]}
when x=0 there is a maximum, so it should be obvious that the tangent is horizontal at `x=0`