graph the function ( y = x ^ { 2 } - 2 x - 8 ) by identifying the domain and any symmetries, finding the…

graph the function ( y = x ^ { 2 } - 2 x - 8 ) by identifying the domain and any symmetries, finding the derivatives ( y ^ { prime } ) and ( y ^ { prime prime } ), 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.\na. the local minimum/minima is/are located at ( ( 1, - 9 ) )\n(type an ordered pair. use a comma to separate answers as needed.)\nb. there are no local minima.\nidentify any local maxima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the local maximum/maxima is/are located at \n(type an ordered pair. use a comma to separate answers as needed.)\nb. there are no local maxima
Answer
Explanation:
Step1: Find the first derivative
Given (y = x^{2}-2x - 8), using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (y^\prime=\frac{d}{dx}(x^{2}-2x - 8)=2x-2).
Step2: Find the critical points
Set (y^\prime = 0), so (2x-2=0). Solving for (x): [ \begin{align*} 2x-2&=0\ 2x&=2\ x&=1 \end{align*} ]
Step3: Find the second derivative
Differentiate (y^\prime = 2x - 2) with respect to (x). Using the power rule, (y^{\prime\prime}=\frac{d}{dx}(2x - 2)=2).
Step4: Analyze the critical point using the second - derivative test
Since (y^{\prime\prime}(1)=2>0), the function has a local minimum at (x = 1). Substitute (x = 1) into the original function (y=x^{2}-2x - 8): (y=(1)^{2}-2(1)-8=1 - 2-8=-9). For local maxima, since (y^{\prime\prime}=2>0) (the function is concave up everywhere, as (y^{\prime\prime}) is a positive constant), there are no points where the function changes from increasing to decreasing (which is required for a local maximum).
Answer:
For local minima: A. The local minimum/minima is/are located at ((1,-9)) For local maxima: B. There are no local maxima