Equation of a tangent line

Question

Answer 1

f(x) = x^3 +x^2 -x-3

has slope = f'(x) = 3x^2+2x-1

f'(-2) = 3(-2)^2 +2(-2)-1 = 12-5=7

f(-2) = -8+4+2-3 = -5

the point of tangency is (-2,-5)

the equation of the tangent line is

y+5 = 7(x+2) in point slope form

or more expressly y - - 5 = 7(x - -2)

or

y = 7x +9 in slope intercept form, with y intercept =9

or

7x-y +9=0 in general form

or

7x-y =-9 in standard form

Answer 2

Determine (-2, f(-2)) this is the point

Determine f'(-2) this is the slope

Use point slope form form Algebra 1.