which equation is y = 9x² + 9x - 1 rewritten in vertex form?\no y = 9(x + 1/2)² - 13/4\no y = 9(x + 1/2)²…

Explanation:

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

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

Step2: Complete the square inside the parentheses

For the quadratic expression $x^{2}+x$, the coefficient of $x$ is $1$. Half of it is $\frac{1}{2}$, and its square is $\frac{1}{4}$. $y = 9(x^{2}+x+\frac{1}{4}-\frac{1}{4})-1$

Step3: Rewrite the expression

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

Step4: Distribute the 9

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

Step5: Simplify the constant terms

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

Answer:

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