Question

find the slope of the graph of the function g(x) = \frac{7x}{x - 3} at (6,14). then find an equation for the line tangent to the graph at that point. the slope of the graph of the function g(x) = \frac{7x}{x - 3} at (6,14) is -\frac{7}{3} (type an integer or a simplified fraction.) the equation for the line tangent to g(x) = \frac{7x}{x - 3} at (6,14) is y = -\frac{7}{3}x + 28.

Explanation:

Step1: Apply quotient - rule for differentiation

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2}$. Here, $u(x) = 7x$, so $u^\prime(x)=7$, and $v(x)=x - 3$, so $v^\prime(x)=1$. Then $g^\prime(x)=\frac{7(x - 3)-7x\times1}{(x - 3)^2}=\frac{7x-21 - 7x}{(x - 3)^2}=-\frac{21}{(x - 3)^2}$.

Step2: Find the slope at the given point

Substitute $x = 6$ into $g^\prime(x)$. $g^\prime(6)=-\frac{21}{(6 - 3)^2}=-\frac{21}{9}=-\frac{7}{3}$.

Step3: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(6,14)$ and $m =-\frac{7}{3}$. So $y-14=-\frac{7}{3}(x - 6)$. Expand the right - hand side: $y-14=-\frac{7}{3}x+14$. Add 14 to both sides to get $y=-\frac{7}{3}x + 28$.

Answer:

The slope of the graph of the function $g(x)=\frac{7x}{x - 3}$ at $(6,14)$ is $-\frac{7}{3}$.
The equation for the line tangent to $g(x)=\frac{7x}{x - 3}$ at $(6,14)$ is $y=-\frac{7}{3}x + 28$.