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How do you find the vertex for #y = 2x^2 + 8x -3#?

Answer 1

Your quadratic is in the form #y=ax^2+bx+c# where: #a=2# #b=8# #c=-3# The coordinates of the vertex are: #x_v=-b/(2a)# #y_v=-Delta/(4a)# Where #Delta=b^2-4ac#

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Answer 2

Julia Adams

To find the vertex of the quadratic function ( y = 2x^2 + 8x - 3 ), you can use the formula for the x-coordinate of the vertex, which is given by:

[ x = \frac{-b}{2a} ]

where ( a ) is the coefficient of the ( x^2 ) term, and ( b ) is the coefficient of the ( x ) term.

In this function, ( a = 2 ) and ( b = 8 ). Substitute these values into the formula:

[ x = \frac{-8}{2(2)} ] [ x = \frac{-8}{4} ] [ x = -2 ]

Once you have found the x-coordinate of the vertex, substitute it back into the original function to find the y-coordinate:

[ y = 2(-2)^2 + 8(-2) - 3 ] [ y = 2(4) - 16 - 3 ] [ y = 8 - 16 - 3 ] [ y = -11 ]

So, the vertex of the quadratic function ( y = 2x^2 + 8x - 3 ) is ( (-2, -11) ).

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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.