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How do you rewrite the equation in vertex form: #4x^2+16-9#?

Answer 1

Ethan Adkins

Version 1: Where I assume an #x# was accidentally missed on the #+16# term: #y=4x^2+16x-9#

Vertex form of a quadratic is #y = m(x-a)^2+b# where the vertex of the parabola is at #(a,b)#

#y=4x^2+16x-9#

#=4(x^2+4x)-9 " extracting the "m" factor"#

#= 4(x^2+4x+2^2) -16 -9" [completing the square](https://tutor.hix.ai)"#

#=4(x+2)-25 " simplifying"#

#=4(x-(-2)) +(-25)" into vertex form"#

(The vertex is at #(x,y) =(-2,-25)#)

Version 2: Where the question was entered correctly (except for the missing #(y=)# which is needed to make it an equation #y=4x^2+16-9# which is equivalent to #y=4x^2+7#

This can be rearranged as #y=4(x-0)^2+7# for a parabola with a vertex at #(x,y)=(0,7)#

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Answer 2

Nathan Axel

To rewrite the equation (4x^2 + 16x - 9) in vertex form, complete the square.

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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.