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: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer