Find the equation of the tangent line of $y=x^2-x$ at $(-2,6)$.

Question

I got the derivative $2x-1$ but I don't know where to go from there. Any help is appreciated.

Answer 1

First, the slope of the tangent is the derivative of the function at point $(x_1, y_1)$, which is said to be $(-2, 6)$. $$f’x = 2x-1$$ $$ m = f’(x_1)$$ $$m = 2\cdot(-2)-1$$ $$\implies \boxed{m = -5}$$ Now, we have the slope, and we have a point. Therefore, we’ll use point-slope form to find the equation of the tangent line. $$y-y_1 = m\cdot(x-x_1)$$ $$y-6 = -5\cdot(x-(-2))$$ $$y-6 = -5\cdot(x+2)$$ $$y-6 = -5x-10 \implies \boxed{y = -5x-4}$$

Answer 2

Try point-slope form: $ y-y_0=m(x-x_0) $

Answer 3

Slope is $2x - 1$ when $x = -2$

Slope $= -4 - 1 = -5$

$y = -5x + b$

$6 = -5(-2) + b$

$b = -4$

Tangent is $y = -5x - 4$