Questions: For the function, find the point(s) on the graph at which the tangent line is horizontal. y=x^3-16x^2+64x+9 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The point(s) at which the tangent line is horizontal is(are) . (Simplify your answer. Type an ordered pair. Use a comma to separate answers as needed.) B. The tangent line is horizontal at all points of the graph. C. There are no points on the graph where the tangent line is horizontal.

Solution Steps

Given the function: \[ y = x^3 - 16x^2 + 64x + 9 \] we first find the derivative to determine the slope of the tangent lines: \[ \frac{dy}{dx} = 3x^2 - 32x + 64 \]

To find where the tangent line is horizontal, we set the derivative equal to zero and solve for \(x\): \[ 3x^2 - 32x + 64 = 0 \] Solving this quadratic equation, we get the critical points: \[ x = \frac{8}{3} \quad \text{and} \quad x = 8 \]

Next, we substitute these \(x\)-values back into the original function to find the corresponding \(y\)-values: \[ y\left(\frac{8}{3}\right) = \left(\frac{8}{3}\right)^3 - 16\left(\frac{8}{3}\right)^2 + 64\left(\frac{8}{3}\right) + 9 = \frac{2291}{27} \approx 84.8519 \] \[ y(8) = 8^3 - 16 \cdot 8^2 + 64 \cdot 8 + 9 = 9 \]

Final Answer