"maximum value of y=x^2+2x+3"?

Question

please help!

as clearly as possible.

thank you!

goshhh i apologize-- this was a typo.

i meant:

maximum value of y=-2x^2+2x+3

THANK YOU!

Answer 1

Since the parabola open up, you don't have the maximum value of y.
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Thank you for correcting the typo.
y = -x^2+2x+3 = -(x-1)^2 + 4
min{y} = 4

Answer 2

Since the parabola open up, you don't have the maximum value of y.
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Thank you for correcting the typo.
y = -x^2+2x+3 = -(x-1)^2 + 4
min{y} = 4

Answer 3

The above poster is correct. And the minimum value, I believe, is minus one (as this is the point at which slope = 0, and the derivative of the formula is 0=2x+2.

Answer 4

x vertex = -b/2a = -2 / 2(2) = -1
x vertex = 1
y vert = x² + 2x + 3
y v = (-1)² + 2(-1) + 3
y v = 1 -2 + 3
y v = 2
or
delta = b² - 4ac = 2² - 4(1)(3) = 4 -12 = -8

y v = -(-8) / 4a = 8/4(1) = 2
:>:

Answer 5

Since this parabola increases positively without bound, there is no maximum.

There is however a minimum value at the point (-1,2).

Answer 6

Maximum value?

If you graph it, it's a parabola. There is no maximum to the y values.

So the answer is infinity.