Question

let $f(x)=\frac{3}{x}$. the slope of the tangent line to the graph of $f(x)$ at the point $(-2,-\frac{3}{2})$ is. the equation of the tangent line to the graph of $f(x)$ at $(-2,-\frac{3}{2})$ is $y = mx + b$ for $m =$ and $b =$. hint: the slope is given by the derivative at $x=-2$, ie. $(lim_{h \to 0}\frac{f(-2 + h)-f(-2)}{h})$. get a common denominator in the numerator and it will then be able to eliminate the h from the denominator.

Explanation:

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

We have $f(x)=\frac{3}{x}=3x^{- 1}$. Using the power - rule for differentiation $\frac{d}{dx}(x^n)=nx^{n - 1}$, we get $f^\prime(x)=-3x^{-2}=-\frac{3}{x^{2}}$.

Step2: Calculate the slope $m$

Substitute $x = - 2$ into $f^\prime(x)$. So $m=f^\prime(-2)=-\frac{3}{(-2)^{2}}=-\frac{3}{4}$.

Step3: Find the value of $b$

We know the equation of the tangent line is $y=mx + b$, and the point $(-2,-\frac{3}{2})$ lies on the line. Substitute $x=-2$, $y =-\frac{3}{2}$ and $m =-\frac{3}{4}$ into $y=mx + b$.
We have $-\frac{3}{2}=-\frac{3}{4}\times(-2)+b$.
Simplify the right - hand side: $-\frac{3}{2}=\frac{3}{2}+b$.
Solve for $b$: $b=-\frac{3}{2}-\frac{3}{2}=-3$.

Answer:

The slope of the tangent line is $-\frac{3}{4}$, $m =-\frac{3}{4}$, $b=-3$.