Question

1. -/3 points how many tangent lines to the curve y = \frac{x}{x + 2} pass through the point (1, 2)? at which exact points do these tangent lines touch the curve? smaller x - value (x, y) = () larger x - value (x, y) = ()

Explanation:

Step1: Find the derivative of the function

First, use the quotient - rule. If $y=\frac{x}{x + 2}=\frac{u}{v}$ where $u=x$ and $v=x + 2$, then $y^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}$. Here, $u^\prime = 1$ and $v^\prime=1$. So $y^\prime=\frac{1\cdot(x + 2)-x\cdot1}{(x + 2)^{2}}=\frac{2}{(x + 2)^{2}}$.
Let the point of tangency on the curve be $(a,\frac{a}{a + 2})$. The slope of the tangent line at $x = a$ is $m=\frac{2}{(a + 2)^{2}}$.

Step2: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$. The tangent line passes through $(a,\frac{a}{a + 2})$ and $(1,2)$. So the slope $m=\frac{\frac{a}{a + 2}-2}{a - 1}$.
Since the slope of the tangent line is also $\frac{2}{(a + 2)^{2}}$, we set up the equation:
$\frac{\frac{a}{a + 2}-2}{a - 1}=\frac{2}{(a + 2)^{2}}$.
Simplify the left - hand side: $\frac{a-2(a + 2)}{(a + 2)(a - 1)}=\frac{2}{(a + 2)^{2}}$.
$\frac{a-2a-4}{(a + 2)(a - 1)}=\frac{2}{(a + 2)^{2}}$.
$\frac{-a - 4}{(a + 2)(a - 1)}=\frac{2}{(a + 2)^{2}}$.
Cross - multiply: $(-a - 4)(a + 2)=2(a - 1)$.
Expand: $-a^{2}-2a-4a-8 = 2a-2$.
Rearrange to get a quadratic equation: $-a^{2}-8a - 6=0$, or $a^{2}+8a + 6=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c=0$ (here $a = 1$, $b = 8$, $c = 6$), use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
$a=\frac{-8\pm\sqrt{8^{2}-4\times1\times6}}{2\times1}=\frac{-8\pm\sqrt{64 - 24}}{2}=\frac{-8\pm\sqrt{40}}{2}=\frac{-8\pm2\sqrt{10}}{2}=-4\pm\sqrt{10}$.

Since the quadratic equation $a^{2}+8a + 6=0$ has two solutions, there are 2 tangent lines.
When $a=-4-\sqrt{10}$:
$y=\frac{-4-\sqrt{10}}{-4-\sqrt{10}+2}=\frac{-4-\sqrt{10}}{-2-\sqrt{10}}=\frac{4+\sqrt{10}}{2+\sqrt{10}}=\frac{(4+\sqrt{10})(\sqrt{10}-2)}{(\sqrt{10}+2)(\sqrt{10}-2)}=\frac{4\sqrt{10}-8 + 10-2\sqrt{10}}{10 - 4}=\frac{2\sqrt{10}+2}{6}=\frac{\sqrt{10}+1}{3}$.
When $a=-4+\sqrt{10}$:
$y=\frac{-4+\sqrt{10}}{-4+\sqrt{10}+2}=\frac{-4+\sqrt{10}}{-2+\sqrt{10}}=\frac{4-\sqrt{10}}{2-\sqrt{10}}=\frac{(4-\sqrt{10})(\sqrt{10}+2)}{(\sqrt{10}-2)(\sqrt{10}+2)}=\frac{4\sqrt{10}+8-10 - 2\sqrt{10}}{10 - 4}=\frac{2\sqrt{10}-2}{6}=\frac{\sqrt{10}-1}{3}$.

Answer:

The number of tangent lines: 2
Smaller $x$-value: $(-4-\sqrt{10},\frac{\sqrt{10}+1}{3})$
Larger $x$-value: $(-4+\sqrt{10},\frac{\sqrt{10}-1}{3})$