Question

For what value of `c` is the line `y=2x + c` a tangent to the parabola `y=x^2-x-2` ?

For what value of `c` is the line `y=2x + c` a tangent to the parabola `y=x^2-x-2` ?

Answer 1

`c= -17/4`

Explanation:

This is an easy calculus problem but it's currently classified as Algebra so let's do it using Algebra I.

`f(x) = x^2 - x - 2`

`f(x) = f(r + (x-r)) = (r+(x-r))^2 - (r+(x-r)) - 2`

`f(x) = r^2 + 2r (x-r) + (x-r)^2 - r - (x-r) - 2`

`f(x) = (r^2 - r - 2) + (2r-1)(x-r) + (x-r)^2`

When `x` is near `r` then `x-r` is small and `(x-r)^2` is smaller still. So the best linear approximation to `f` near `x=r` is gotten by dropping the `(x-r)^2` term. That's the tangent line:

`y = (r^2 - r - 2) + (2r-1) (x-r)`

`y = (2r-1)x -r^2-2`

Comparing that to `y=2x+c` we get

`2r - 1 = 2`

`r = 3/2`

`c = -r^2 - 2 = - 9/4 - 2 = - 17/4`

Check: Plot (2x - 17/4 - y)(x^2 - x - 2 - y) = 0

graph{(2x - 17/4 - y)(x^2 - x - 2 - y) = 0 [-3, 5, -3, 2]}