Question

How do you find the equations of the two tangent lines to the graph of `f(x) = x^2` that pass through the point (-1, -8)?

Answer 1

the two equations are: `y=4x-4` and `y=-8x-16`

Explanation:

`f(x)=x^2 => f'(x)=2x`

A generic point on the curve can be represented by `(a,f(a))`, ie `(a,a^2)`, and the gradient of the tangent at that point has gradient `m=f'(a)`, or `m=2a`,

Hence using `y-y_1=m(x-x_1)` we can find the equation of the tangent at a generic point `(a,a^2)`:
`y-a^2 = (2a)(x-a)`
`:. y-a^2 = 2ax-2a^2`
`:. y = 2ax-a^2`

If this tangent passes through `(-1,-8)` then:
`:. -8 = 2a(-1)-a^2`
`:. -8 = -2a-a^2`
`:. a^2+2a-8=0`

This is a quadratic in a (which is good as we expected to find two equation, and therefore we require two values of `a`!)
`a^2+2a-8 = 0`
`:. (a+4)(a-2) = 0`
`:. a=2,a=-4`

When `a=-4 => y=2(-4)x-(-4)^2`
`:. y=-8x-16`

When `a=2 => y=2(2)x-(2)^2`
`:. y=4x-4`

So the two equations are: `y=4x-4` and `y=-8x-16`