Please help me with these questions. It's my test and I don't know how to do it.

I did #2, you can do #1

2a) The parabola opens upward because the a coefficient (in front of x^2) is positive

2b) There is a minimum point of (3,-2) which is where the vertex is. You can find the x-value of the vertex using -b/2a and the y-value by substituting the x-value. However, there is no maximum point since the parabola opens upward infinitely.

2c) The axis of symmetry is x = 3 because it is x = -b/2a or -(-12)/2(2) = 12/4 = 3

2d) The vertex is (3,-2) since the x-intercept of the vertex is 3 (given from the equation x = -b/2a). You get 3 and when you substitute x as 3 into the parabolas equation (2*3^2 - 12(3) +16) you get y = -2.

2e) To find the x-intercepts, factor the equation:

2(x^2 - 6x + 8)

We need two numbers to add to -6 and multiply to 8

These two numbers are -4 and -2

y = 2(x-4)(x-2)

The x-intercepts are where y = 0

0 = 2(x-4)(x-2)

0 = x - 4; x = 4

0 = x - 2; x = 2

the x-intercept are (4, 0) and (2, 0)

2f) The y-intercept is when x = 0 or the c value of the equation (the constant)

2(0)^2 - 12(0) + 16 = 0 - 0 + 16 = 16

The y-intercept is (0, 16)

2g) The domain is All Real Numbers

2h) The range is y >= -2

2i) It is increasing when x > 3

2j) It is decreasing when x < 3

2k) I can't graph here but some points on the graph are (3,-2) , (0,16) , (1,6) , (5, 6)

there are no answer choices

step-by-step explanation:

answer: your correct answer is b. d.

step-by-step explanation: