Questions: For f(x)=2x^4-16x^2+6 find the following. (A) f'(x) (B) The slope of the graph of f at x=-3 (C) The equation of the tangent line at x=-3 (D) The value(s) of x where the tangent line is horizontal (A) f'(x)=8x^3-32x (B) At x=-3, the slope of the graph of f is.

For f(x)=2x^4-16x^2+6 find the following.
(A) f'(x)
(B) The slope of the graph of f at x=-3
(C) The equation of the tangent line at x=-3
(D) The value(s) of x where the tangent line is horizontal
(A) f'(x)=8x^3-32x
(B) At x=-3, the slope of the graph of f is.

Transcript text: For $f(x)=2 x^{4}-16 x^{2}+6$ find the following. (A) $f^{\prime}(x)$ (B) The slope of the graph of $f$ at $x=-3$ (C) The equation of the tangent line at $x=-3$ (D) The value(s) of $x$ where the tangent line is horizontal (A) $f^{\prime}(x)=8 x^{3}-32 x$ (B) At $\mathrm{x}=-3$, the slope of the graph of f is $\square$.

Solution

Solution Steps

Step 1: Find the Derivative

To find $f^{\prime}(x)$ , we differentiate the function $f(x) = 2x^4 - 16x^2 + 6$ . The derivative is given by: $f^{\prime}(x) = 8x^3 - 32x$

Step 2: Evaluate the Slope at

x = -3

Next, we evaluate the derivative at $x = -3$ to find the slope of the graph at that point: $f^{\prime}(-3) = -120$

Step 3: Find the Equation of the Tangent Line at

x = -3

To find the equation of the tangent line at $x = -3$ , we first calculate the $y$ -coordinate of the function at that point: $f(-3) = 24$ Using the point-slope form of the line, the equation of the tangent line is: $y - 24 = -120(x + 3)$ This simplifies to: $y = -120x - 336$