What is the vertex form of $y = 3 x^{2} + 9 x - 6$ $y= 3x^2 + 9x-6$ ?

Question

3108 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-02T12:59:01

$3 {(x - \frac{3}{2})}^{2} - \frac{51}{4}$ $3(x-3/2)^2-51/4$

The vertex form of the quadratic expression $a x^{2} + b x + c$ $ax^2+bx+c$ is $a {(x - h)}^{2} + k$ $a(x-h)^2+k$ .

We have $3 x^{2} + 9 x - 6$ $3x^2+9x-6$

Or:

$3 (x^{2} + 3 x) - 6$ $3(x^2+3x)-6$

To complete the square, we take the coefficient of $3 x$ $3x$ , divide it by $2$ $2$ , square it and add it:

$3 (x^{2} + 3 x + \frac{9}{4}) - 6 - \frac{27}{4}$ $3(x^2+3x+9/4)-6-27/4$

$3 {(x - \frac{3}{2})}^{2} - \frac{51}{4}$ $3(x-3/2)^2-51/4$

So the vertex is, incidentally, $(\frac{3}{2}, - \frac{51}{4})$ $(3/2,-51/4)$

What is the vertex form of y=3x2+9x−6y= 3x^2 + 9x-6?