What is the vertex of #y= -7(2x-1)^2-3#?

Answer 1

Julia Adams

The vertex is #(1/2,-3)#

The vertex form of quadratic function is

#y=a(x-h)^2+k#

Where #(h,k)# is the vertex.

Our problem is #y=-7(2x-1)^2-3#

Let us try to convert this to the form #y=a(x-h)^2+k#

#y=-7(2(x-1/2))^2 -3 #

#y=-7(2^2)(x-1/2)^2-3#

#y=-7(4)(x-1/2)^2 - 3#

#y=-28(x-1/2)^2 - 3#

Now comparing with #y=a(x-h)^2 +k#

We can see #h=1/2# and #k=-3#

The vertex is #(1/2,-3)#

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Answer 2

Nathan Axel

#Vertex (1/2, -3)#

This is actually the vertex form of y. x-coordinate of vertex: (2x - 1) = 0 --> #x = 1/2# y-coordinate of vertex: y = -3

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Answer 3

Abigail Allen

To find the vertex of the quadratic equation y = -7(2x - 1)^2 - 3, we use the standard form y = a(x - h)^2 + k. In this form, the vertex is at the point (h, k). For the given equation, h = 1 and k = -3. So, the vertex is (1, -3).

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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.