graph the function y = x² - 2x - 8 by identifying the domain and any symmetries, finding the derivatives y…

graph the function y = x² - 2x - 8 by identifying the domain and any symmetries, finding the derivatives y and y, finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. find the second derivative y. y = identify any critical points. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) occur(s) at x = . (use a comma to separate answers as needed.) b. there are no critical points. identify any local minima. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the local minimum/minima is/are located at . (type an ordered pair. use a comma to separate answers as needed.) b. there are no local minima.
Answer
Explanation:
Step1: Find the first - derivative
Given $y = x^{2}-2x - 8$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y'=\frac{d}{dx}(x^{2}-2x - 8)=2x-2$.
Step2: Find the second - derivative
Differentiate $y' = 2x - 2$ with respect to $x$. Using the power rule again, $y''=\frac{d}{dx}(2x - 2)=2$.
Step3: Find critical points
Set $y'=0$. So, $2x - 2=0$. Solving for $x$ gives $2x=2$, then $x = 1$.
Step4: Determine local minima
Since $y''=2>0$, the function is concave up at $x = 1$. Substitute $x = 1$ into the original function $y=(1)^{2}-2(1)-8=1 - 2-8=-9$. So the local minimum is at the point $(1,-9)$.
Answer:
$y'' = 2$ A. The critical point(s) occur(s) at $x = 1$ A. The local minimum/minima is/are located at $(1,-9)$