Question

At what point on the quadratic y=`x^2`-4x-1 is the tangent parallel to the straight line y=-2x+5?

Answer 1

`(1, -4)`

Explanation:

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

`d/(dx) (-2x+5) = -2`

So we are trying to find where these two slopes are equal, that is:

`2x - 4 = -2`

Hence `x = 1`, then `x^2-4x-1 = -4`

graph{(y-x^2+3.92x+1)(y+2x-4.9)(y+2x+2)((x-1)^2+(y+4)^2-0.05) = 0 [-4.95, 9.29, -6.04, 1.076]}