How do you write $y=x^2-2x-9$ in vertex form?

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Answer 1 · 2017-07-01T17:42:12

Please see the explanation.

Given: $y=x^2-2x-9" [1]"$

We observe that equation [1] is in the standard form:

$y = ax^2+bx+c" [2]"$

where $a = 1, b = -2, and c = -9$

The vertex form of this type of parabola is:

$y = a(x-h)^2+k" [3]"$

The "a" in equation [2] and the "a" in equation [3] are the same attribute of a parabola, therefore, we may substitute 1 for "a" into equation [3]:

$y = (x-h)^2+k" [4]"$

We know that "h" is the x coordinate of the axis of the vertex given by the formula:

$h = (-b)/(2a)$

Substituting in the know values:

$h = (-(-2))/(2(1))$

$h = 1$

Substitute the value of h into equation [4]:

$y = (x-1)^2+k" [5]"$

We know that k is the y coordinate of the vertex. We can find the value of k by evaluating the function at h:

$k = 1^2-2(1) -9$

$k = -10$

Substitute the value of k into equation [5]:

$y = (x-1)^2-10" [6]"$

Equation [6] is the vertex form.

How do you write y=x^2-2x-9y=x^2-2x-9 in vertex form?