For what values of x is the tangent line of the graph of
f(x)=(6x^3)+(36x^2)-36
parallel to the line
y=−1.4 ?
To find the values of x for which the tangent line of the graph of f(x) = 6x^3 + 36x^2 - 36 is parallel to the line y = -1.4, we need to find the derivative of f(x) and then determine when the derivative is equal to the slope of the given line.
Step 1: Find the derivative of f(x)
Taking the derivative of f(x), we get:
f'(x) = 3(6x^2) + 2(36x) = 18x^2 + 72x
Step 2: Determine the slope of the given line
The given line y = -1.4 is in the form y = mx + b, where m is the slope. In this case, the slope is -1.4, since it is the coefficient of x.
Step 3: Equate the derivative of f(x) to the slope of the line
Set 18x^2 + 72x equal to -1.4:
18x^2 + 72x = -1.4
Step 4: Solve the equation for x
To solve this quadratic equation, we'll set it equal to zero and use the quadratic formula:
18x^2 + 72x + 1.4 = 0
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), where a = 18, b = 72, and c = 1.4:
x = (-72 ± √(72^2 - 4(18)(1.4))) / (2(18))
x = (-72 ± √(5184 - 100.8)) / 36
x = (-72 ± √(5083.2)) / 36
Step 5: Simplify and approximate the values of x
Using a calculator, we find:
x ≈ (-72 ± 71.34) / 36
Simplifying further, we get two possible values of x:
x ≈ -2.495 or x ≈ -0.005
Therefore, for values of x approximately equal to -2.495 or -0.005, the tangent line of the graph of f(x) is parallel to the line y = -1.4.