Question

The tangent to the curve $y=x^2e^x$ at $(2,4e^2)$ also passes through the point.

• A. ($\frac{1}{2},0$)
• B. ($\frac{3}{2},0$)
• C. (0,1)
• D. ($\frac{5}{2},0$)

Answer 1

Answer: (B)

($\frac{3}{2},0$)

Answer 2

To find the point through which the tangent to the curve $y=x^2e^x$ at $(2,4e^2)$ also passes, we follow these steps:

1. Find the derivative of the curve:
   Given the curve $y = x^2e^x$.
   Using the product rule, $\frac{dy}{dx} = \frac{d}{dx}(x^2)e^x + x^2\frac{d}{dx}(e^x)$.
   $\frac{dy}{dx} = 2xe^x + x^2e^x$
   Factor out $e^x$:
   $\frac{dy}{dx} = e^x(2x + x^2)$

2. Calculate the slope of the tangent at the given point $(2, 4e^2)$:
   Substitute $x=2$ into the derivative:
   $m = \left(\frac{dy}{dx}\right)_{x=2} = e^2(2(2) + 2^2)$
   $m = e^2(4 + 4)$
   $m = 8e^2$

3. Determine the equation of the tangent line:
   The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.
   Here, $(x_1, y_1) = (2, 4e^2)$ and $m = 8e^2$.
   So, the equation of the tangent line is:
   $y - 4e^2 = 8e^2(x - 2)$
   $y - 4e^2 = 8e^2x - 16e^2$
   $y = 8e^2x - 16e^2 + 4e^2$
   $y = 8e^2x - 12e^2$

4. Check which of the given options satisfies the tangent line equation:

   • Option 1: $(\frac{1}{2},0)$
     Substitute $x=\frac{1}{2}$, $y=0$:
     $0 = 8e^2(\frac{1}{2}) - 12e^2$
     $0 = 4e^2 - 12e^2$
     $0 = -8e^2$ (False)

   • Option 2: $(\frac{3}{2},0)$
     Substitute $x=\frac{3}{2}$, $y=0$:
     $0 = 8e^2(\frac{3}{2}) - 12e^2$
     $0 = 12e^2 - 12e^2$
     $0 = 0$ (True)

Since this option satisfies the equation, it is the correct answer.