Question

Which equation is y = 9x2 + 9x – 1 rewritten in vertex form?

Answer 1

The vertex form of a parabolic function has the general formula:

f(x) = a(x-h)^2 + k where (h,k) represent the vertex of the parabola.

Therefore, to write the given equation in vertex form, we will need to transform it to the above formula as follows:

y = 9x^2 + 9x - 1

y = 9(x^2 + x) - 1

y = 9(x^2 + x + 1/4 - 1/4)-1

y = 9((x+1/2)^2 - 1/4)-1

y = 9(x + 1/2)^2 - 9/4 - 1

y = 9(x + 1/2)^2 - 13/4 ..............> The equation in vertex form

Explanation:

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Answer 2

`y = 9(x +(1)/(2)) ^ 2 -(13)/(4)`

Explanation:

An equation in the vertex form is written as

`y = a (x-h) + k`

Where the point (h, k) is the vertex of the equation.

For an equation in the form
`ax ^ 2 + bx + c` the x coordinate of the vertex is defined as

`x = -(b)/(2a)`

In this case we have the equation
`y = 9x^2 + 9x - 1`.

Where

`a = 9\\\\b = 9\\\\c = -1`

Then the x coordinate of the vertex is:

`x = -(9)/(2(9))\\\\x = -(9)/(18)\\\\x = -(1)/(2)`

The y coordinate of the vertex is replacing the value of
`x = -(1)/(2)` in the function

`y = 9 (-0.5) ^ 2 + 9 (-0.5) -1\\\\y = -(13)/(4)`

Then the vertex is:

`(-(1)/(2), -(13)/(4))`

Therefore The encuacion excrita in the form of vertice is:

`y = a(x +(1)/(2)) ^ 2 -(13)/(4)`

To find the coefficient a we substitute a point that belongs to the function
`y = 9x^2 + 9x - 1`

The point (0, -1) belongs to the function. Thus.

`-1 = a(0 + (1)/(2)) ^ 2 -(13)/(4)`

`-1 = a((1)/(4)) -(13)/(4)\\\\a = (-1 +(13)/(4))/((1)/(4))\\\\a = 9`

Then the written function in the form of vertice is

`y = 9(x +(1)/(2)) ^ 2 -(13)/(4)`