Question 252705: determine the vertex of quadratic function stating axis of symmetry,minimum or maximum value,and domain and range on y=10x^2+5x-1
and y=-24x^2+18x+11
I am not understanding even though i read over examples.
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! let's start with the first equation and then move to the second:
quadratics are expressed as Y = Ax^2 + Bx + C.
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vertex:
We can find the vertex using -b/2a. -5/2*10 = -5/20 = -1/4. This is the x part of the vertex and now we put that into f(x) to get
10(1/4)^2 + 5(1/4)-1 = 10/16 + 5/4 - 1 = 14/16 = 7/8.
vertex = (1/4,7/8)
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axis of symmetry:
This is just the x value of the vertex
x = 1/4.
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min:
since the graph opens up, there will be a min at the vertex.
min @ (1/4,7/8)
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domain:domain: For a parabola opening up or down, domain will be all reals.
range: Since the parabola opens up, the minimum range is 7/8. So we have Y >= 7/8.
-- EQUATION #2
{{y = -24x^2 + 18x + 11}}
quadratics are expressed as Y = Ax^2 + Bx + C.
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vertex:
We can find the vertex using -b/2a. -18/2*-24 = 18/48 = 3/8. This is the x part of the vertex and now we put that into f(x) to get
-24(3/8)^2 + 18(3/8) + 11 = -27/8 + 27/4 + 11 = 115/4.
vertex = (3/8, 115/4)
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axis of symmetry:
This is just the x value of the vertex
x = 3/8.
--
min: No min in this case.
max: SInce the graph opens down, there will be a max at the vertex.
min @ (3/8, 115/4)
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domain: For a parabola opening up or down, domain will be all reals.
range: Since the parabola opens down, the highest range is 115/4. So we have Y <= 115/4.
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