Answer
$(a)$
$x$-intercept is $A(-1,0)$
$y$-intercept is $B(0,1)$
$(b)$
$x$ and $y$-intercepts are $A(0,0)$. The graph is symmetrical about $y$-axis.
See the images below.
Work Step by Step
$(a)$ $y=\sqrt{x+1}$
We have $x$-intercept when $y=0$
$0=\sqrt{x+1}$
$x+1=0$
$x=-1$
$x$-intercept, $A(-1,0)$
$y$-intercept $x=0$
$y=sqrt{1}$
$y=1$
$y$-intercept is $B(0,1)$
The graph has a form of $y=\sqrt{x}$. We can sketch the graph by approximately connecting critical points. For a better visualization, let's find one more random point.
$x=3$
$y=\sqrt{3+1}$
$y=2$
$C(3,2)$
See the graph above. It has no symmetry.
$(b)$ $y=-|x|$
We have $x$-intercept when $y=0$
$0=-|x|$
$x=0$
$x$-intercept, $A(0,0)$
For $y$-intercept $x=0$
$y=-|0|$
$y=0$
$y$-intercept is $A(0,0)$
We have a graph with form of $y=|x|$, but it is reflected about $x$-axis. See the graph below.
The graph is symmetric about $y$-axis.
