mth 261 module 1 exam (sec 1.1 - 1.8) - form 1-21-a (honorlock) started: apr 6 at 6:36pm quiz instructions…
mth 261 module 1 exam (sec 1.1 - 1.8) - form 1-21-a (honorlock) started: apr 6 at 6:36pm quiz instructions show instructions question 8 6 pts find an equation for the line tangent to the graph of the given function at the indicated point. f(x)=x² - x at (-2, 6) edit view insert format tools table 12pt paragraph b i u a t²
Answer
Explanation:
Step1: Find the derivative of the function
The derivative of $f(x)=x^{2}-x$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=2x-1$.
Step2: Evaluate the derivative at the given x - value
Substitute $x = - 2$ into $f'(x)$. So $f'(-2)=2(-2)-1=-4 - 1=-5$. The slope of the tangent line $m=-5$.
Step3: Use the point - slope form of a line
The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-2,6)$ and $m=-5$. Substitute these values: $y - 6=-5(x+2)$.
Step4: Simplify the equation
Expand the right - hand side: $y - 6=-5x-10$. Add 6 to both sides to get the equation in slope - intercept form: $y=-5x - 4$.
Answer:
$y=-5x - 4$