which equation is y = 9x² + 9x - 1 rewritten in vertex form?\n○ y = 9(x + 1/2)² - 13/4\n○ y = 9(x + 1/2)²…

Explanation:

Step1: Factor out the coefficient of $x^{2}$ from the first - two terms

$y = 9x^{2}+9x - 1=9(x^{2}+x)-1$

Step2: Complete the square inside the parentheses

For the quadratic expression $x^{2}+x$, we know that $(x + a)^{2}=x^{2}+2ax + a^{2}$. In $x^{2}+x$, if $2a = 1$, then $a=\frac{1}{2}$ and $x^{2}+x=(x+\frac{1}{2})^{2}-\frac{1}{4}$. So $y = 9((x+\frac{1}{2})^{2}-\frac{1}{4})-1$

Step3: Expand the expression

$y=9(x + \frac{1}{2})^{2}-\frac{9}{4}-1$

Step4: Simplify the constant terms

$y=9(x+\frac{1}{2})^{2}-\frac{9 + 4}{4}=9(x+\frac{1}{2})^{2}-\frac{13}{4}$

Answer:

$y = 9(x+\frac{1}{2})^{2}-\frac{13}{4}$ (It seems there is a mistake in the provided options as the correct vertex - form we derived is not among them. But the process to convert to vertex - form is as above.)