$\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
y=a(x-(-4))^2-1\implies y=a(x+4)^2-1
\\\\\\
\textit{now, we also know that }
\begin{cases}
y=0\\
x=2
\end{cases}\implies \underline{0}=a(\underline{2}+4)^2-1$

solve for "a"

5 0

3 years ago

Solve for x xx. Your answer must be simplified. x + 5 ≤ − 4

Rama09 [41]

Answer:

x ≤ − 9

Step-by-step explanation:

x + 5 ≤ − 4

Subtract 5 from each side

x + 5-5 ≤ − 4-5

x ≤ − 9

8 0

3 years ago

The first equation from the previous system of

VladimirAG [237]

Answer: (3, -2)

Step-by-step explanation:

The solution to the system is where the graphs intersect.

3 0

2 years ago

I don’t really understand this

frozen [14]

Use a calculator and multiple

4 0

3 years ago