Answer:
a) The slope of the secant line is 11 + h
b) The slope of the curve at P is 11
c) The equation of the tangent line at P is y = 11x - 11
Explanation:
y=x²+3x+5
(a) Find the slope of the secant line to the curve through the points P=(4,33) and Q=(4+h,(4+h)²+3(4+h)+5)
(4+h)²+3(4+h)+5 = 16 + 8h + h² + 12 + 3h + 5 = 33 + 11h + h²
slope (m) = yq - yp/xq - xp → m = 33 + 11h + h² - 33/4 + h - 4
m = 11h + h²/h = h(11+h)/h = 11 + h
The slope of the secant line is 11 + h
(b) Use your answer from part (a) to find the slope of the curve at the point P.
As P (4,33) = (4+0, 33+0) at P h is 0, so the slope at P is 11 + 0 = 11
The slope of the curve at P is 11
(c) Write an equation of the tangent line to the curve at P.
y - yp = m(x - xp) → y - 33 = 11(x - 4) → y - 33 = 11x - 44 → y = 11x - 11
The equation of the tangent line at P is y = 11x - 11