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How do you write #y = x^2 - 9x - 10# into vertex form?

Answer 1

Nathan Axel

#(y+121/4)=(x-9/2)^2#

#" "y=x^2-9x-10#

#" "y=x^2-2*9/2x+81/4-121/4#

#" "y+121/4=x^2-2*9/2x+81/4#

#" "(y+121/4)=(x-9/2)^2#

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Answer 2

Kennedy Adams

To write the quadratic function ( y = x^2 - 9x - 10 ) into vertex form, follow these steps:

Complete the square to factor the quadratic.
Rewrite the quadratic in vertex form by completing the square.

Completing the square involves adding and subtracting a constant term to the expression in a way that allows it to be written as the square of a binomial.

Complete the square: [ y = x^2 - 9x - 10 ] [ y = (x^2 - 9x) - 10 ] To complete the square, add and subtract ((\frac{-9}{2})^2 = (\frac{-9}{2})^2 = \frac{81}{4}): [ y = (x^2 - 9x + \frac{81}{4}) - \frac{81}{4} - 10 ] [ y = (x^2 - 9x + \frac{81}{4}) - \frac{41}{4} ]
Rewrite in vertex form: [ y = (x - \frac{9}{2})^2 - \frac{41}{4} ]

So, the quadratic function ( y = x^2 - 9x - 10 ) in vertex form is: [ y = (x - \frac{9}{2})^2 - \frac{41}{4} ]

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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.