Question

at what point(s) on the graph of $f(x)=x^3 + 12x^2 + 54x + 14$ is the slope of the tangent line 6?

the slope of the tangent line is 6 at $\square$.
(simplify your answer. type an ordered pair. use a comma to separate answers as needed.)

Explanation:

Step1: Find derivative of $f(x)$

The derivative $f'(x)$ gives the slope of the tangent line. Using power rule:
$f'(x) = 3x^2 + 24x + 54$

Step2: Set slope equal to 6

Set $f'(x)=6$ and simplify:
$3x^2 + 24x + 54 = 6$
$3x^2 + 24x + 48 = 0$
Divide by 3: $x^2 + 8x + 16 = 0$

Step3: Solve quadratic equation

Factor the quadratic:
$(x+4)^2 = 0$
$x = -4$

Step4: Find corresponding $y$-value

Substitute $x=-4$ into $f(x)$:
$f(-4) = (-4)^3 + 12(-4)^2 + 54(-4) + 14$
$f(-4) = -64 + 192 - 216 + 14 = -74$

Answer:

$(-4, -74)$