y = 4 x^2 - 5 x
y' = 8 x - 5
at x = 2
y'(2) = 16-5 = 11
m = 2
y = m x + b
6 = 2 (2) + b
b = 2
so
y = 2 x + 2
f(x) = 4x2 - 5x
(a) Find f '(2).
(b) Use the answer in part (a) to find an equation of the tangent line to the parabola y = 4x2 - 5x at the point (2, 6).
y =
y' = 8 x - 5
at x = 2
y'(2) = 16-5 = 11
m = 2
y = m x + b
6 = 2 (2) + b
b = 2
so
y = 2 x + 2
(a)
To find f'(x), we will differentiate each term of the function separately. Applying the power rule, the derivative of 4x^2 is 8x, and the derivative of -5x is -5. Therefore, f'(x) = 8x - 5.
To find f'(2), we substitute x = 2 into the derivative function f'(x):
f'(2) = 8(2) - 5
= 16 - 5
= 11
Therefore, f'(2) = 11.
(b)
To find the equation of the tangent line to the parabola at the point (2, 6), we use the point-slope form of a linear equation: y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line.
Since f'(2) gives us the slope of the tangent line, and the given point (2, 6) lies on the line, we substitute m = 11, x1 = 2, and y1 = 6 into the point-slope form equation:
y - 6 = 11(x - 2)
Simplifying the equation, we get:
y - 6 = 11x - 22
Finally, we isolate y to get the equation of the tangent line:
y = 11x - 22 + 6
y = 11x - 16
Therefore, the equation of the tangent line to the parabola y = 4x^2 - 5x at the point (2, 6) is y = 11x - 16.