Calculus- find an equation of the tangent line to the curve

Question

If f(x) = 5x^2 − x^3, find f ′(1) and use it to find an equation of the tangent line to the curve y = 5x^2 − x^3 at the point (1, 4).

Answer 1

f '(x) = (2)(5)x^2-1-(3)(x^3-1) = 10x - 3x² , f '(1) = 10(1)-3(1)² = 7 Please note that f '(1) is also the slope of tangent line to the curve at x =1, the slope is therefore = 7 a positive increasing slope. The equation of the tangent line at the point ( 1 , 4) is a line that passes through the point (1 ,4 ) on the curve that is f(1) = 5(1)² - 3(1)³ = 4

writing the equation in slope form m = y-y1/ x-x1, and where m=7, it becomes 7 = y-4/x-1, rearranging,

y-4 = 7(x-1), y-4 = 7x -7, and y = 7x-7+4, y = 7x -3 is the equation of the tangent line. If this is written in y intercept form y = mx + b, m =7, and b = -3

To determine if the line passes through the point (1,4) substitute x =1, y = 7(1) - 3 = 4 and the line passes through the point (1,4). If we substitute x = 0, y = 7(0)-3 = -3 and the line also passes through the point (0,-3) which is the y intercept point

Answer 2

f'(x) = y' = 10x-3x^2

f'(1) = y'(1) = 10-3 = 7

tangent line is

y=mx + b where m=slope = 7 and b = y intercept. Plug in the point (1,4) to solve for b

4= 7 +b

b =-3

y=7x-3 is the tangent line to the cubic equation at the point (1,4)