Answer to Question #312318 in Calculus for shh

Question

Find the equation of the tangent to at the point where , in f(x)=x^2+4x-5

standard form.

Answer 1

We proceed to differentiate: f'(x)=2x+4

Then we substitute for the point x₀:

$y-y_0=f'(x_0)*(x-x_0) \\ y-y_0=(2x_0+4)*(x-x_0) \\ y=y_0+(2x)x_0-2x_0^2+4x-4x_0 \\ y=x_0^2+4x_0-5+(2x)x_0-2x_0^2+4x-4x_0 \\ y=2x(x_0+2)-x_0^2-5=mx+b$

This leads us to the conclusion that the standard form for the tangent line is the line with slope $m=2(x_0+2)$ and the intersection at $b=-(x_0^2+5)$, where x₀ is the point where the tangent line is placed.