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: \[ y = -120x - 336 \]