What is the vertex form of $y= 6x^2 - 4x - 24$ ?

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Answer 1 · 2016-07-02T17:08:24

$y = 6(x-1/3)^2 - 24 2/3$

The vertex is at $(1/3 . -24 2/3)$

If you write a quadratic in the form

$a(x +b)^2 +c$ , then the vertex is $(-b,c)$

Use the process of completing the square to get this form:

$y = 6x^2 - 4x -24$

Factor out the 6 to make $6x^2$ into $"x^2$

$y = 6(x^2 -(2x)/3 - 4)" " 4/6 = 2/3$

Find half of $2/3$ ................................. $2/3 ÷ 2 = 1/3$

square it....... $(1/3)^2$ and add it and subtract it.

$y = 6[x^2 -(2x)/3 color(red)(+ (1/3)^2) - 4 color(red)(- (1/3)^2)]$

Write the first 3 terms as the square of a binomial

$y = 6[(x-1/3)^2 - 4 1/9]$

Multiply the 6 into the bracket to get the vertex form.

$y = 6(x-1/3)^2 - 24 2/3$

The vertex is at $(1/3 . -24 2/3)$

What is the vertex form of y= 6x^2 - 4x - 24 y= 6x^2 - 4x - 24 ?