Question

Find the slope of the tangent line to the graph of fat the given point.$f\left(x\right)=5-2x,at(-3,11)$.

Short answer

The slope of the tangent line is $-2$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$\begin{array}{l}f\left(x\right)=5-2x\\ po\mathrm{int}=(-3,11)\end{array}$

Step 2. Formula used.

The tangent line to the curve$y=f\left(x\right)$at the point$P(a,f(a\left)\right)$is the line through p with slope,

$m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

provided that this limit exists.

Step 3. Finding the slope of tangent line at given point.

Let$f\left(x\right)=5-2x$,then the slope of the tangent line at$(-3,11)$is,

$\begin{array}{c}m=\underset{x\to -3}{\lim }\frac{f\left(x\right)-f(-3)}{x-(-3)}\\ =\underset{x\to -3}{\lim }\frac{(5-2x)-(5-2\cdot (-3\left)\right)}{x+3}\\ =\underset{x\to -3}{\lim }\frac{5-2x-5-6}{x+3}\\ =\underset{x\to -3}{\lim }\frac{-2x-6}{x+3}\\ =\underset{x\to -3}{\lim }\frac{-2\cancel{(x+3)}}{\cancel{(x+3)}}\\ =-2\end{array}$

Therefore, the slope of the tangent line is$-2$ .