Question

1. find the real roots for $y = x^2 + 6x + 9$. use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

options: (3,0), (-3,0), (-3,-3), (3,3)

1. find the vertex for $y = x^2 + 6x + 9$. use $\left(\frac{-b}{2a}, \frac{-(b^2 - 4ac)}{4a}\

ight)$ from the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
options: (3,0), (-3,0), (-3,-3), (-3,3)

Explanation:

Step1: Identify a, b, c for Q8

For $y = x^2 + 6x + 9$, $a=1$, $b=6$, $c=9$

Step2: Substitute into quadratic formula

$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(9)}}{2(1)}$

Step3: Calculate discriminant

$\sqrt{36 - 36} = \sqrt{0} = 0$

Step4: Solve for x

$x = \frac{-6 \pm 0}{2} = -3$

Step5: Identify a, b, c for Q9

For $y = x^2 + 6x + 9$, $a=1$, $b=6$, $c=9$

Step6: Calculate x-coordinate of vertex

$x = \frac{-b}{2a} = \frac{-6}{2(1)} = -3$

Step7: Calculate y-coordinate of vertex

$y = \frac{-(b^2 - 4ac)}{4a} = \frac{-(36 - 36)}{4(1)} = 0$

Answer:

1. $\boldsymbol{(-3, 0)}$
2. $\boldsymbol{(-3, 0)}$