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What is the vertex form of #y= (x-1)(x – 6) #?

Answer 1

Eva Abramson

#y=(x-7/2)^2 -25/4#

To solve for the vertex form, we can "complete the square" after converting this to standard form.

#y=(x-1)(x-6)#

#y=x^2-6x-x+6#

#y=x^2-7x+6#

Now let's complete the square. To do that, we need to find a value that make #x^2-7x# a perfect square. To find that value, we take the middle term, #-7#, and we divide it by #2#. That gives us #-7/2#. Now we square the fraction: #49/4#

Now we have the value that makes the equation true. But!! we cannot introduce a new value! Not without immediately subtracting it, which would make the final value #0#.

#y=x^2-7x+6 + 49/4 - 49/4#

So, we added #49/4# and then #-49/4#. Now let's rearrange it so we have a perfect square.... and other stuff:

#y=x^2-7x+49/4+6-49/4#

Let's rewrite #x^2-7x+49/4# as a perfect square: #(x-7/2)^2#

Now our equation is #y=(x-7/2)^2 +6-49/4#

mix similar terms

#y=(x-7/2)^2 -25/4#

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Answer 2

Isaiah Anthony

The vertex form of the given quadratic function y = (x - 1)(x - 6) is y = x^2 - 7x + 6.

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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.