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What is the vertex of # y=1/5(x/2-15)^2-2 #?

Answer 1

vertex: #(30,-2)#

Our "target will be to convert the given equation into "vertex form": #color(white)("XXX")y=m(x-color(red)(a))^2+color(blue)(b)# with vertex at #(color(red)(a),color(blue)(b))#

Given #color(white)("XXX")y=1/2(x/2-15)^2-2#

#y=1/2((x-30)/2)^2-2#

#y=1/2(((x-30)^2)/(2^2))-2#

#y=1/8(x-color(red)(30))^2+color(blue)("("-2")")#

which is the vertex form with a vertex at #(color(red)(30),color(blue)(-2))#

Our answer is (at least roughly) correct, as shown by the graph below: graph{1/5(x/2-15)^2-2 [9.41, 49.99, -10.61, 9.69]}

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Answer 2

Isaiah Anthony

The vertex of the given quadratic function y = (1/5)(x/2 - 15)^2 - 2 is (30, -2).

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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.