SOLUTION: Can someone please help me understand this question? I would like to know the process of how you come to the answers as I have a test tomorrow! y=-2(x-3)^2+8 a)State the coordina

Question 704340: Can someone please help me understand this question? I would like to know the process of how you come to the answers as I have a test tomorrow!
y=-2(x-3)^2+8
a)State the coordinates of the vertex
b)State whether is has a max or min value and what that is?
c)State the equation of the axis of symmetry
d)State the y-intercept.
If you could show me how you reach each answer..that would be amazing!
Thank you
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
y=-2(x-3)^2+8
a)State the coordinates of the vertex
You have to remember the "vertex form" of a quadratic:
y = a(x-h)^2 + k
where (h,k) is the vertex
since your equation was:
y=-2(x-3)^2+8
we immediately can see that the vertex is at:
(h,k) = (3,8)
b)State whether is has a max or min value and what that is?
We can tell by the "coefficient" associated with the x^2 term:
y=-2(x-3)^2+8
expanding the above:
y=-2(x^2-6x+9)+8
y=-2x^2+12x-18+8
y=-2x^2+12x-10
Since it will be "negative" -- we conclude that it will be a parabola that opens downwards. Thus the vertex will be the MAX.
Note: do part c first, then come back to this
c)State the equation of the axis of symmetry
axis of symmetry is at:
x = -b/(2a)
x = -12/(2(-2))
x = -12/(-4)
x = 3 (axis of symmetry)
d)State the y-intercept.
set x to zero:
y=-2x^2+12x-10
y=-2(0)^2+12(0)-10
y=0+0-10
y = -10
y-intercept is at (0, -10)

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