Question

find an equation of the tangent line to the graph of $f(x)=\frac{x + 3}{x^{2}+2}$ at $x = 1$. options: $5x+9y = 7$, $9y - 5x=7$, $5x + 9y=17$, $5x+3y = 9$

Explanation:

Step1: Find the derivative of $f(x)$

Use the quotient - rule. If $f(x)=\frac{u(x)}{v(x)}$ where $u(x)=x + 3$ and $v(x)=x^{2}+2$, then $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^{2}}$. Here, $u^\prime(x) = 1$ and $v^\prime(x)=2x$. So $f^\prime(x)=\frac{(x^{2}+2)-2x(x + 3)}{(x^{2}+2)^{2}}=\frac{x^{2}+2-2x^{2}-6x}{(x^{2}+2)^{2}}=\frac{-x^{2}-6x + 2}{(x^{2}+2)^{2}}$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f^\prime(x)$. $f^\prime(1)=\frac{-1^{2}-6\times1 + 2}{(1^{2}+2)^{2}}=\frac{-1-6 + 2}{9}=\frac{-5}{9}$.

Step3: Find the value of $y$ at $x = 1$

Substitute $x = 1$ into $f(x)$. $f(1)=\frac{1 + 3}{1^{2}+2}=\frac{4}{3}$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,\frac{4}{3})$ and $m=-\frac{5}{9}$. So $y-\frac{4}{3}=-\frac{5}{9}(x - 1)$. Multiply through by 9 to get $9y-12=-5(x - 1)$. Expand to $9y-12=-5x + 5$. Rearrange to $5x+9y=17$.

Answer:

$5x + 9y=17$