SOLUTION: Ok, I'm not sure if this problem is in this section but I didn't see a "parabolas" title anywhere. I know the answer to this problem, but I have know idea how to get it. It's fin

Question 57124This question is from textbook College Algebra: Graphs and Models
: Ok, I'm not sure if this problem is in this section but I didn't see a "parabolas" title anywhere. I know the answer to this problem, but I have know idea how to get it. It's finding the y-coordinate of the vertex of this parabola function:
g(x)= 2x^2+6x+8
I know how to find the x-coordinate, and when I checked it, I was right, but my answer for the y-coordinate is 7/4 and the book says its supposed to be 7/2. Please help me, I have a test on this tomorrow and I really need to know what I'm leaving out of this problem so I don't do it on the test!!! This question is from textbook College Algebra: Graphs and Models

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
It's finding the y-coordinate of the vertex of this parabola function:
g(x)= 2x^2+6x+8
I know how to find the x-coordinate, and when I checked it, I was right, but my answer for the y-coordinate is 7/4 and the book says its supposed to be 7/2
:
Let's just go through the problem: Vertex equation: x = -b/(2a)
In your equation, b = 6 and a = 2
:
x = -6/(2*2)
x = -6/4 reduces to -3/2
:
Find the y coordinate, substitute -3/2 for x in the original equation:
y = 2(-3/2)^2 + 6(-3/2) + 8
y = 2(9/4) - 18/2 + 8
y = 18/4 - 9 + 8
Y - 18/4 - 1
y = 18/4 - 4/4
y = 14/4 reduces to 7/2
:
did this help?
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