Using the completing-the-square method, find the vertex of the function f(x) = 2x2 − 8x + 6 and indicate whether it is a minimum or a maximum and at what point. maximum at (2, –2) minimum at (2, –2) maximum at (2, 6) minimum at (2, 6)

  Minimum at (2, -2)

Step-by-step explanation:

Factor the leading coefficient from the first two terms:

  f(x) = 2(x^2 -4x) +6

Inside parentheses, add the square of half the x-coefficient. Add the opposite of that amount outside parentheses.

  f(x) = 2(x^2 -4x +4) +6 -8

Rewrite in vertex form.

  f(x) = 2(x -2)^2 -2

The positive leading coefficient tells you the vertex is a minimum. Comparing this to the vertex form for a quadratic with vertex (h, k)

  f(x) = a(x -h)^2 +k

we find the vertex to be ...

  (h, k) = (2, -2)

The function has a minimum at (2, -2).

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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