Question

in exercises 11 and 12, graph g. 11. $f(x)=3(x + 1)^2-1;g(x)=f(x + 2)$ 12. $f(x)=\frac{1}{2}(x - 3)^2-5;g(x)=-f(x)$ vertex: a.o.s.: y - int: vertex: a.o.s.: y - int: 264 algebra 1 student journal copyright © big ideas learning, llc all rights reserved.

Explanation:

Step1: Analyze Exercise 11

Given $f(x)=3(x + 1)^2-1$ and $g(x)=f(x + 2)$. First, substitute $x$ with $x+2$ in $f(x)$:
$g(x)=3((x + 2)+1)^2-1=3(x + 3)^2-1$.
For a quadratic function in vertex - form $y=a(x - h)^2+k$, the vertex is $(h,k)$. Here, for $g(x)=3(x + 3)^2-1$, $h=-3$ and $k=-1$, so the vertex is $(-3,-1)$. The axis of symmetry (A.O.S.) is $x = h=-3$. To find the $y$-intercept, set $x = 0$: $g(0)=3(0 + 3)^2-1=3\times9-1=27-1 = 26$.

Step2: Analyze Exercise 12

Given $f(x)=\frac{1}{2}(x - 3)^2-5$ and $g(x)=-f(x)$. Then $g(x)=-\frac{1}{2}(x - 3)^2 + 5$. For the quadratic function $g(x)=-\frac{1}{2}(x - 3)^2+5$, the vertex is $(3,5)$ (since for $y=a(x - h)^2+k$, $h = 3$ and $k = 5$). The axis of symmetry is $x=h = 3$. To find the $y$-intercept, set $x = 0$: $g(0)=-\frac{1}{2}(0 - 3)^2+5=-\frac{1}{2}\times9 + 5=-\frac{9}{2}+5=\frac{1}{2}$.

Answer:

For Exercise 11:
Vertex: $(-3,-1)$
A.O.S.: $x=-3$
$y$-int: $26$
For Exercise 12:
Vertex: $(3,5)$
A.O.S.: $x = 3$
$y$-int: $\frac{1}{2}$