Question

find the equation of the line tangent to the given curve at a. use a graphing utility to graph the curve and the tangent line on the same set of axes. y = \frac{6x}{x + 2}; a = 4. the equation for the tangent line is y = . (type an expression using x as the variable.)

Explanation:

Step1: Find the derivative using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 6x$, $u^\prime=6$, $v=x + 2$, $v^\prime = 1$. So $y^\prime=\frac{6(x + 2)-6x\times1}{(x + 2)^{2}}=\frac{6x+12 - 6x}{(x + 2)^{2}}=\frac{12}{(x + 2)^{2}}$.

Step2: Find the slope of the tangent line at $x = a$

Substitute $x = 4$ into the derivative. $m=y^\prime|_{x = 4}=\frac{12}{(4 + 2)^{2}}=\frac{12}{36}=\frac{1}{3}$.

Step3: Find the y - coordinate at $x = a$

Substitute $x = 4$ into the original function $y=\frac{6x}{x + 2}$. $y=\frac{6\times4}{4+2}=\frac{24}{6}=4$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,4)$ and $m=\frac{1}{3}$. So $y-4=\frac{1}{3}(x - 4)$.

Step5: Simplify the equation

$y-4=\frac{1}{3}x-\frac{4}{3}$, then $y=\frac{1}{3}x-\frac{4}{3}+4=\frac{1}{3}x+\frac{8}{3}$.

Answer:

$y=\frac{1}{3}x+\frac{8}{3}$