How do you write the Vertex form equation of the parabola #f(x) = x^2 - 6x + 5#?

Answer 1

#y = (x - 3 )^2 - 4#

The standard form of a quadratic function is y = #ax^2 + bx + c#
the function here #y = x^2 - 6x + 5 " is in this form"#

in contrast, a = 1, b = - 6, and c = 5.

The vertex form of the parabola is # y = a(x - h )^2 + k # where ( h , k ) are the coords of the vertex.
x-coord of vertex # =( -b)/(2a) = -(-6)/2 = 3 #
and y-coord = #(3)^2 -6(3) + 5 = 9 - 18 + 5 = -4 #

Consequently, a = 1 and (h, k) = (3, -4)

#rArr y = (x - 3 )^2 - 4 " is equation in vertex form "#
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Answer 2

To write the vertex form equation of the parabola ( f(x) = x^2 - 6x + 5 ), complete the square. The vertex form is ( f(x) = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. Completing the square for the given equation yields ( f(x) = (x - 3)^2 - 4 ). Therefore, the vertex form equation is ( f(x) = (x - 3)^2 - 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.

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.

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.

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.

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