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How do you find the maximum value of #y=x^2+2x+3#?

Answer 1

There is no maximum value.

The graph is an upward opening parabola; there is no maximum.

2) #lim_(xrarroo)(x^2+2x+3) = oo#, so there is no maximum.

Using #y = f(x)#,

3) #f'(x) = 2x+2 = 0# at #x=-1#. Therefore, #-1# is the only critical number and the first or second derivative test will show that #f(-1)# is a minimum, not a maximum.

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Answer 2

Luke Alvarez

To find the maximum value of ( y = x^2 + 2x + 3 ), you need to determine the vertex of the parabola represented by this quadratic function. The vertex of a parabola given by the equation ( y = ax^2 + bx + c ) is located at the point ((h, k)), where ( h = -\frac{b}{2a} ) and ( k = f(h) ) (the value of the function at ( h )).

For ( y = x^2 + 2x + 3 ), ( a = 1 ) and ( b = 2 ). So, ( h = -\frac{2}{2(1)} = -1 ).

Now, substitute ( x = -1 ) into the equation to find ( k ): ( k = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2 ).

Therefore, the maximum value of ( y ) is ( k = 2 ), which occurs at ( x = -1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.