61 - 64. equations of tangent lines
a. find an equation of the line tangent to the given curve at a.
b. use a graphing utility to graph the curve and the tangent line on the same set of axes.
61. $y=\frac{x + 5}{x - 1};a = 3$
62. $y=\frac{2x^{2}}{3x - 1};a = 1$
63. $y = 1+2x+xe^{x};a = 0$
64. $y=\frac{e^{x}}{x};a = 1$

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Accepted Answer

1. **For problem 61: $y=\frac{x + 5}{x - 1}$, $a = 3$**
   - **Step - 1: Find the derivative of $y$ using the quotient - rule**
     - The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u=x + 5$, so $u^\prime=1$, and $v=x - 1$, so $v^\prime=1$.
     - $y^\prime=\frac{1\cdot(x - 1)-(x + 5)\cdot1}{(x - 1)^{2}}=\frac{x - 1-x - 5}{(x - 1)^{2}}=\frac{-6}{(x - 1)^{2}}$.
   - **Step - 2: Find the slope of the tangent line at $x = a=3$**
     - Substitute $x = 3$ into $y^\prime$: $y^\prime(3)=\frac{-6}{(3 - 1)^{2}}=\frac{-6}{4}=-\frac{3}{2}$.
   - **Step - 3: Find the $y$ - coordinate when $x = 3$**
     - Substitute $x = 3$ into $y=\frac{x + 5}{x - 1}$: $y(3)=\frac{3 + 5}{3 - 1}=\frac{8}{2}=4$.
   - **Step - 4: Use the point - slope form $y - y_1=m(x - x_1)$ to find the equation of the tangent line**
     - Here, $x_1 = 3$, $y_1 = 4$, and $m=-\frac{3}{2}$.
     - $y - 4=-\frac{3}{2}(x - 3)$.
     - Expand: $y-4=-\frac{3}{2}x+\frac{9}{2}$.
     - Rearrange to the slope - intercept form $y=-\frac{3}{2}x+\frac{9}{2}+4=-\frac{3}{2}x+\frac{9 + 8}{2}=-\frac{3}{2}x+\frac{17}{2}$.
2. **For problem 62: $y=\frac{2x^{2}}{3x - 1}$, $a = 1$**
   - **Step - 1: Find the derivative of $y$ using the quotient - rule**
     - Let $u = 2x^{2}$, then $u^\prime=4x$, and $v=3x - 1$, then $v^\prime=3$.
     - $y^\prime=\frac{4x(3x - 1)-2x^{2}\cdot3}{(3x - 1)^{2}}=\frac{12x^{2}-4x - 6x^{2}}{(3x - 1)^{2}}=\frac{6x^{2}-4x}{(3x - 1)^{2}}$.
   - **Step - 2: Find the slope of the tangent line at $x = a = 1$**
     - Substitute $x = 1$ into $y^\prime$: $y^\prime(1)=\frac{6\times1^{2}-4\times1}{(3\times1 - 1)^{2}}=\frac{6 - 4}{4}=\frac{1}{2}$.
   - **Step - 3: Find the $y$ - coordinate when $x = 1$**
     - Substitute $x = 1$ into $y=\frac{2x^{2}}{3x - 1}$: $y(1)=\frac{2\times1^{2}}{3\times1 - 1}=\frac{2}{2}=1$.
   - **Step - 4: Use the point - slope form $y - y_1=m(x - x_1)$ to find the equation of the tangent line**
     - Here, $x_1 = 1$, $y_1 = 1$, and $m=\frac{1}{2}$.
     - $y - 1=\frac{1}{2}(x - 1)$.
     - Expand: $y-1=\frac{1}{2}x-\frac{1}{2}$.
     - Rearrange to the slope - intercept form $y=\frac{1}{2}x-\frac{1}{2}+1=\frac{1}{2}x+\frac{1}{2}$.
3. **For problem 63: $y=1 + 2x+xe^{x}$, $a = 0$**
   - **Step - 1: Find the derivative of $y$ using the sum - rule and product - rule**
     - The sum - rule: $(u + v+w)^\prime=u^\prime+v^\prime+w^\prime$. Let $u = 1$, $v = 2x$, $w=xe^{x}$.
     - $u^\prime = 0$, $v^\prime=2$. For $w = xe^{x}$, use the product - rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = x$, $u^\prime = 1$, $v = e^{x}$, $v^\prime=e^{x}$, so $w^\prime=e^{x}+xe^{x}$.
     - $y^\prime=0 + 2+(e^{x}+xe^{x})=2+e^{x}(1 + x)$.
   - **Step - 2: Find the slope of the tangent line at $x = a = 0$**
     - Substitute $x = 0$ into $y^\prime$: $y^\prime(0)=2+e^{0}(1 + 0)=2 + 1=3$.
   - **Step - 3: Find the $y$ - coordinate when $x = 0$**
     - Substitute $x = 0$ into $y=1 + 2x+xe^{x}$: $y(0)=1+2\times0+0\times e^{0}=1$.
   - **Step - 4: Use the point - slope form $y - y_1=m(x - x_1)$ to find the equation of the tangent line**
     - Here, $x_1 = 0$, $y_1 = 1$, and $m = 3$.
     - $y - 1=3(x - 0)$.
     - So $y=3x + 1$.
4. **For problem 64: $y=\frac{e^{x}}{x}$, $a = 1$**
   - **Step - 1: Find the derivative of $y$ using the quotient - rule**
     - Let $u = e^{x}$, then $u^\prime=e^{x}$, and $v = x$, then $v^\prime=1$.
     - $y^\prime=\frac{e^{x}\cdot x-e^{x}\cdot1}{x^{2}}=\frac{e^{x}(x - 1)}{x^{2}}$.
   - **Step - 2: Find the slope of the tangent line at $x = a = 1$**
     - Substitute $x = 1$ into $y^\prime$: $y^\prime(1)=\frac{e^{1}(1 - 1)}{1^{2}}=0$.
   - **Step - 3: Find the $y$ - coordinate when $x = 1$**
     - Substitute $x = 1$ into $y=\frac{e^{x}}{x}$: $y(1)=\frac{e^{1}}{1}=e$.
   - **Step - 4: Use the point - slope form $y - y_1=m(x - x_1)$ to find the equation of the tangent line**
     - Here, $x_1 = 1$, $y_1 = e$, and $m = 0$.
     - $y - e=0(x - 1)$.
     - So $y = e$.

# Answer:
- For problem 61: The equation of the tangent line is $y=-\frac{3}{2}x+\frac{17}{2}$.
- For problem 62: The equation of the tangent line is $y=\frac{1}{2}x+\frac{1}{2}$.
- For problem 63: The equation of the tangent line is $y=3x + 1$.
- For problem 64: The equation of the tangent line is $y = e$.

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