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

graph the function ( y = x^{2}-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\nc. the function has no oblique asymptotes.\ngraph the function ( y = x^{2}-2x - 8 ). choose the correct graph below.\na.\nb.\nc.\nd.

graph the function ( y = x^{2}-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\nc. the function has no oblique asymptotes.\ngraph the function ( y = x^{2}-2x - 8 ). choose the correct graph below.\na.\nb.\nc.\nd.

Answer

Explanation:

Step1: Determine the domain

The function ( y = x^{2}-2x - 8 ) is a polynomial function. Polynomial functions have a domain of all real numbers, ( (-\infty,\infty) ).

Step2: Check for symmetries

For a function ( y = f(x) ), if ( f(-x)=f(x) ) (even function, symmetric about the ( y -)axis) and ( f(-x)=-f(x) ) (odd function, symmetric about the origin). ( f(-x)=(-x)^{2}-2(-x)-8=x^{2}+2x - 8\neq x^{2}-2x - 8=f(x) ) and ( f(-x)\neq-(x^{2}-2x - 8) ). The axis of symmetry of a parabola ( y = ax^{2}+bx + c ) is given by ( x=-\frac{b}{2a} ). Here ( a = 1), ( b=-2 ), so ( x =-\frac{-2}{2\times1}=1 ).

Step3: Find the first - derivative

Using the power rule ( (x^{n})^\prime=nx^{n - 1} ), ( y^\prime=\frac{d}{dx}(x^{2}-2x - 8)=2x-2 ). Set ( y^\prime = 0) (critical points): ( 2x-2=0\Rightarrow2x=2\Rightarrow x = 1 ). When ( x\lt1), let ( x = 0), then ( y^\prime(0)=2\times0 - 2=-2\lt0 ), so the function is decreasing on the interval ( (-\infty,1) ). When ( x\gt1), let ( x = 2), then ( y^\prime(2)=2\times2 - 2=2\gt0 ), so the function is increasing on the interval ( (1,\infty) ). The function has a local minimum at ( x = 1). Substitute ( x = 1) into ( y=x^{2}-2x - 8): ( y=(1)^{2}-2(1)-8=1 - 2 - 8=-9 ).

Step4: Find the second - derivative

( y^{\prime\prime}=\frac{d}{dx}(2x - 2)=2). Since ( y^{\prime\prime}=2\gt0) for all ( x\in(-\infty,\infty) ), the function is concave up on ( (-\infty,\infty) ) and there are no inflection points (because the concavity does not change).

Step5: Find the ( x -)intercepts

Set ( y = 0), ( x^{2}-2x - 8=0). Factor: ( (x - 4)(x+2)=0). So ( x=-2) or ( x = 4). Set ( x = 0) to find the ( y -)intercept: ( y=(0)^{2}-2(0)-8=-8 ).

Step6: Analyze asymptotes

Since ( y=x^{2}-2x - 8) is a polynomial function (degree ( n = 2)), there are no vertical or horizontal asymptotes. And as ( n = 2\gt1), there are no oblique asymptotes.

For a parabola ( y=ax^{2}+bx + c) with ( a = 1\gt0), it opens upwards. The vertex (local minimum) is at ( (1,-9)), ( x -)intercepts at ( (-2,0)) and ( (4,0)), ( y -)intercept at ( (0,-8)).

Answer:

The correct graph is the one that is a parabola opening upwards (since (a = 1>0)) with vertex at ((1,-9)), (x -)intercepts at ((-2,0)) and ((4,0)), and (y -)intercept at ((0,-8)). If we assume the options:

  • Option A: parabola opening downwards (incorrect as (a>0))
  • Option B: parabola opening upwards with vertex ((1,-9)), (x=-2) and (x = 4) as roots (correct)
  • Option C: (incorrect if it does not match the intercepts and vertex)
  • Option D: parabola opening downwards (incorrect as (a>0))

So the answer is B.