11. -/2 points details my notes find equations of the tangent line and normal line to the given curve at the…

Explanation:

Step1: Find the derivative of $y = 7xe^{x}$

Use the product - rule $(uv)^\prime=u^\prime v + uv^\prime$, where $u = 7x$ and $v=e^{x}$. Then $u^\prime = 7$ and $v^\prime=e^{x}$. So $y^\prime=7e^{x}+7xe^{x}=7e^{x}(1 + x)$.

Step2: Find the slope of the tangent line at $(0,0)$

Substitute $x = 0$ into $y^\prime$. When $x = 0$, $y^\prime(0)=7e^{0}(1 + 0)=7$.

Step3: Find the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, with $(x_{1},y_{1})=(0,0)$ and $m = 7$. So $y-0 = 7(x - 0)$, which simplifies to $y = 7x$.

Step4: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is $7$, the slope of the normal line is $-\frac{1}{7}$.

Step5: Find the equation of the normal line

Use the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(0,0)$ and $m=-\frac{1}{7}$. So $y-0=-\frac{1}{7}(x - 0)$, which simplifies to $y=-\frac{1}{7}x$.

Answer:

tangent line: $y = 7x$ normal line: $y=-\frac{1}{7}x$