Question

let $f(x)=2x^{2}-6x + 13$. the slope of the tangent line to the graph of $f(x)$ at the point $(2,9)$ is $square$. the equation of the tangent line to the graph of $f(x)$ at $(2,9)$ is $y = mx + b$ for $m=square$ and $b=square$. hint: the slope is given by the derivative at $x = 2$, ie. $lim_{x
ightarrow2}\frac{f(2 + h)-f(2)}{h}$

Explanation:

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

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=2x^{2}-6x + 13$, then $f'(x)=4x-6$.

Step2: Find the slope $m$

Substitute $x = 2$ into $f'(x)$. So $m=f'(2)=4\times2-6=8 - 6=2$.

Step3: Find the value of $b$

We know the equation of the line is $y=mx + b$, and the line passes through the point $(2,9)$. Substitute $x = 2$, $y = 9$ and $m = 2$ into $y=mx + b$. We get $9=2\times2 + b$. Then $9 = 4 + b$, and $b=9 - 4=5$.

Answer:

The slope of the tangent line to the graph of $f(x)$ at the point $(2,9)$ is $2$.
$m = 2$
$b = 5$