Question

For the function f(x) = –2(x + 3)2 − 1, identify the vertex, domain, and range.

Answer 1

\[f(x) = –2(x + 3)^2 − 1\]
domain ; no denominator ,no radical ---->\[D_f=R\]

range:
\[f(x) = –2(x + 3)^2 − 1\\y=–2(x + 3)^2 − 1\\y+1=-2(x+3)^2\\(x+3)^2=\frac{y+1}{-2}\\(x+3)^2 \geq 0 \rightarrow \frac{y+1}{-2} \geq 0\\ \rightarrow y+1 \leq 0 \rightarrow y \leq -1\]


for vertex
\[y=a(x+b)^2 +c \rightarrow x_{vertex}=-b\]

Answer 2

so would the vertex be (-3, -1) or (3, -1) ?

Answer 3

-3 right?

Answer 4

okay yeah i got it right thank you!