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What is the vertex of #y= 1/4(x+2)^2 - 9#?

Answer 1

#(-2, -9)#

This problem is actually already set up in vertex form. From here, we have all the info we need. #1/4(xcolor(green)(+)color(blue)(2))^2color(red)(-9)# tells us that the vertex is #(color(green)(-)color(blue)(2), color(red)(-9))#.

Notice that the sign switched for #color(blue)(2)#. But That's the only really "tricky" thing about thid type of problem. It Really is pretty easy. Just switch the sign for the #color(blue)(x)#-component, and leav the sign alone for the #color(red)(y)#-component.

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Answer 2

Ethan Adkins

The vertex of the parabola given by the equation ( y = \frac{1}{4}(x + 2)^2 - 9 ) is (-2, -9).

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Answer 3

Ethan Adkins

The vertex of the quadratic function ( y = \frac{1}{4}(x+2)^2 - 9 ) can be found using the vertex form of a quadratic function, which is ( y = a(x-h)^2 + k ), where ((h, k)) represents the vertex of the parabola.

Comparing the given function with the vertex form, we see that ( h = -2 ) and ( k = -9 ).

Therefore, the vertex of the function ( y = \frac{1}{4}(x+2)^2 - 9 ) is ((-2, -9)).

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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.