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How do you find the vertex of #y=2x^2 +11x-6#?

Answer 1

So, the solution set - or vertex set - is #(-11/4,-169/8)#

The x of the vertex(#x-vertex#) is found using the formula: #"x-vertex" = -b /(2a) #, in this case: # "x-vertex" = -11/4#. The y of the vertex(#y-vertex#) is found using the formula: #"y-vertex" = - triangle/(4a) = - (b^2 -4ac) / (4a)#, in this case: #"y-vertex" = - [121 - (4 " x " 2" x"-6)] / (4 " x "2) =- 169/8 #. So, the solution set - or vertex set - is #(-11/4,-169/8)#

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Answer 2

Savannah Anderson

To find the vertex of the quadratic function y = 2x^2 + 11x - 6, you can use the formula for the x-coordinate of the vertex, which is given by:

x = -b / (2a)

where 'a' is the coefficient of the x^2 term (2 in this case), and 'b' is the coefficient of the x term (11 in this case).

Substituting the values, you get:

x = -11 / (2 * 2)

Simplify to get:

x = -11 / 4

Once you have the x-coordinate of the vertex, you can find the corresponding y-coordinate by substituting this value back into the original equation.

Then, y = 2(-11/4)^2 + 11(-11/4) - 6

Calculate to find the y-coordinate.

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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.