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What is the vertex form of #y=7x^2-9x-32#?

Answer 1

Ethan Adkins

#y_("vertex form")=7(x-9/14)^2-977/28#

Given: #y=7x^2-9x-32#......................(1)

Write as:

#y=7(x^2-9/7x)-32#

Now write as

#y=7(x-[1/2xx9/7])^2-32 color(blue)(+"correction")#

#y=7(x-9/14)^2-32color(blue)(+"correction")#..........................(2) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider #7(x-9/14)^2# This gives: #7(x^2-9/7x+81/196)# We need the #7(x^2-9/7x)# but the #7(+81/196)# is an extra value we need to get rid of. This is why we have a correction. In this case the correction value is:#color(blue)(7( -81/196)=-81/28)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So equation (2) becomes:

#y=7(x-9/14)^2-32color(blue)(+(-81/28))#..........................(2_a)

#y=7(x-9/14)^2-977/28#

#("So " x_("vertex")=(-1)xx(-9/14)color(white)(..)=+9/14)#

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Answer 2

Eva Adamson

The vertex form of the quadratic equation y = 7x^2 - 9x - 32 is y = 7(x - (9/14))^2 - 457/28.

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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.