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What is the vertex form of #y=x^2-4x+9#?

Answer 1

Savannah Anderson

#y-5=(x-2)^2#

The equation of the parabola is

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

#rArr" "y-5=x^2-4x+4#

#rArr" "y-5=(x-2)^2#

So the parabola is of the form #x^2=4ay# with the vertex at #(2,5)#.

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Answer 2

Eva Adamson

The vertex form of the quadratic equation (y = ax^2 + bx + c) is (y = a(x - h)^2 + k), where (h, k) is the vertex of the parabola. To find the vertex form of (y = x^2 - 4x + 9), first, complete the square for the quadratic expression (x^2 - 4x) by adding and subtracting the square of half the coefficient of x, which is ((-4/2)^2 = 4):

(y = x^2 - 4x + 9 - 9).

Now, rewrite the first three terms as a squared binomial:

(y = (x^2 - 4x + 4) - 9).

This simplifies to:

(y = (x - 2)^2 - 9).

Therefore, the vertex form of (y = x^2 - 4x + 9) is (y = (x - 2)^2 - 9).

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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.