Question 1047662: Absolutely stumped. What is the max or min value of (x) = 2x^2-3x+5.
Worked out the Y intercept for a value of (0,5), and the vertex, of (0.75, 3.875). Cannot for the life of me work out how to get a number that isn't imaginary for the x intercepts.
Found 2 solutions by Alan3354, ewatrrr:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Absolutely stumped. What is the max or min value of (x) = 2x^2-3x+5.
Worked out the Y intercept for a value of (0,5), and the vertex, of (0.75, 3.875). Cannot for the life of me work out how to get a number that isn't imaginary for the x intercepts.
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Maybe it is imaginary (or complex).
check the discriminant:
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Disc = b^2 - 4ac = 9 - 4*2*5 = -31
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Disc < 0 --> no real solutions, no x-intercepts.
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But the problem says find the max or min.
The coefficient of x^2 is >0 --> it's a minimum
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The min is on the AOS, Axis of Symmetry, which is x = -b/2a
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x = 3/4
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f(3/2) = 2(9/16) - 3(3/4) + 5 = 9/8 - 18/8 + 40/8 = 31/8 --- I didn't want to take your word for it.
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The vertex is the minimum. The fact that there is no x-intercept is not relevant.
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It's a minimum of 31/8.
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Re: TY
Know what You mean. Making a sketch would often help me to get the answer, as well.
y = 2x^2-3x+5 Parabola Opening Upward
V(.75,3.875) Yes
min for y is 3.875
x-intercept Arithmetically 'silly' because there is none
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