Question

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

a. the absolute maximum value occurs at ( x = )
(use a comma to separate answers as needed. type each answer only once.)
b. there is no absolute maximum.
identify the absolute minimum value and where it occurs. select the correct choice below and,
if necessary, fill in the answer boxes to complete your choice.
a. the absolute minimum value occurs at ( x = )
(use a comma to separate answers as needed. type each answer only once.)
b. there is no absolute minimum.

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 \(x\), we get \(2x=2\), then \(x = 1\).

Step3: Find the second derivative

Differentiate \(y^\prime=2x - 2\) with respect to \(x\). Using the power rule, \(y^{\prime\prime}=\frac{d}{dx}(2x - 2)=2\).
Since \(y^{\prime\prime}=2>0\), the function is concave - up everywhere.

Step4: Analyze the function at the critical point

Substitute \(x = 1\) into the original function \(y=x^{2}-2x - 8\). Then \(y=(1)^{2}-2\times(1)-8=1 - 2-8=-9\).
As \(y^{\prime\prime}>0\), the function has a local minimum at \(x = 1\). For a quadratic function \(y=ax^{2}+bx + c\) (\(a = 1>0\)), the parabola opens upwards. So the local minimum is also the absolute minimum.

Answer:

A. The absolute minimum value \(-9\) occurs at \(x = 1\).