Calculus Derivatives

Question

Find the y-intercept of the tangent line at the point P (-2, f(-2)) on the graph of

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

Answer 1

P(-2, f(-2))

f(-2) = (-2)²+ 2(-2) + 3

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

P(-2, 3)

f(x) = x²+ 2x + 3

f’(x) = 2x + 2

f’(-2) = 2(-2) + 2

f’(-2) = -4 + 2 = -2

y = -2x + b

3 = -2(-4) + b

3 = 8 + b

b = -5

y-intercept = -5