Question

let $f(x)=2x^{2}-10x + 10$. the slope of the tangent line to the graph of $f(x)$ at the point $(3,-2)$ is . the equation of the tangent line to the graph of $f(x)$ at $(3,-2)$ is $y = mx + b$ for $m = $ and $b= $. hint: the slope is given by the derivative at $x = 3$. question help: video

Explanation:

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

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=2x^{2}-10x + 10$, then $f'(x)=4x-10$.

Step2: Calculate the slope $m$

Substitute $x = 3$ into $f'(x)$. So $m=f'(3)=4\times3-10=12 - 10=2$.

Step3: Find the $y$-intercept $b$

We know the equation of the line is $y=mx + b$, and the line passes through the point $(3,-2)$. Substitute $x = 3$, $y=-2$ and $m = 2$ into $y=mx + b$. We get $-2=2\times3 + b$. Then $b=-2-6=-8$.

Answer:

The slope of the tangent line at the point $(3,-2)$ is $2$.
$m = 2$
$b=-8$