Question

Find the slope and the equation of the tangent of the curve at the given point. Show your solution.

1) y = 2x² -x + 2 (-1, 5)

2) y = x² + x + 1 at (-2, 3)

Answer 1

equation of the tangent of the curve at the point a can be found the following way

$f(x)-f(a)=f'(a)(x-a)$ , f'(a) is the slope of this tangent line

1) y = 2x² -x + 2 (-1,5)

$f(-1)=5$

$f'(x)=4x-1\implies f'(-1)=-5$ - slope

So, the sought equation is $f(x)-5=-5(x+1)\implies f(x)=-5x$

2) y = x² + x + 1 at (-2, 3)

$f(-2)=3$

$f'(x)=2x+1\implies f'(-2)=-3$ - slope

So, the sought equation is $f(x)-3=-3(x+2)\implies f(x)=-3x-3$