Question

For what values of x is the tangent line of the graph

of
f(x)= 6.8x^3+33.66x^2-40.76x+4.08
parallel to the line y=1-2x?
x=____

Answer 1

x = -3.8 or x = 0.5

Step-by-step explanation:

f(x) = 6.8x³ + 33.66x² - 40.76x + 4.08

f'(x) = 20.4x² + 67.32x - 40.76

y = 1 - 2x; m = -2

f'(x) = 20.4x² + 67.32x - 40.76 = -2

20.4x² + 67.32x - 40.76 = -2

20.4x² + 67.32x - 38.76 = 0

10x² + 33x - 19 = 0

(5x + 19)(2x - 1) = 0

5x + 19 = 0 or 2x - 1 = 0

5x = -19 or 2x = 1

x = -3.8 or x = 0.5

Answer 2

The values of x are found by setting the derivative of f(x) equal to the slope of y=1-2x which is -2, and solving the resulting quadratic equation.

Step-by-step explanation:

To find the values of x for which the tangent line to the graph of f(x) = 6.8x^3 + 33.66x^2 - 40.76x + 4.08 is parallel to the line y = 1 - 2x, we need to determine when the derivative of f(x) is equal to the slope of the given line, which is -2. The derivative of f(x) is f'(x) = 20.4x^2 + 67.32x - 40.76. Setting this equal to -2 gives us a quadratic equation to solve:

20.4x^2 + 67.32x - 40.76 = -2

20.4x^2 + 67.32x - 38.76 = 0

We can solve this quadratic equation using the quadratic formula to find the values of x. The solutions to this equation will give us the desired x-values for which the tangent lines are parallel to y = 1 - 2x.