Answer
$(a)$
$x$-intercept $A(3,0)$
$y$-intercept $B(0,-6)$
$(b)$
$x$-intercept $B(-1,0)$
$y$-intercept $A(0,-1)$
Also see the images below.
Work Step by Step
$(a)$ To graph the given linear function, we need at least $2$ points corresponding this function. But, for a better visualization we will take $3$ points, $x$-intercept, $y$-intercept and a point corresponding to a random $x$ value.
$2x-y=6$
We have $x$-intercept when $y=0$
$2x-0=6$
$2x=6$
$x=3$
We have first point, $x$-intercept, $B(3,0)$
For $y$-intercept $x=0$
$2\times0-y=6$
$-y=6$
$y=-6$
$y$-intercept is $A(0,-6)$
Now let's write a function of $y$ in terms of $x$, then take any real number and calculate $3^{rd}$ point (Take different value than the intercepts we found earlier).
$2x-y=6$
$y=2x-6$
Let $x$ be $2$
$y=2\times2-6$
$y=-2$
We found $C(2,-2)$
To sketch the graph, we simply plot these points and connect them. Also see the image above.
The graph is symmetric to only a line perpendicular to this line.
$(b)$ $y=-(x+1)^2$
We have $x$-intercept when $y=0$
$0=-(x+1)^2$
$x+1=0$
$x=-1$
We have first point, $x$-intercept, $B(-1,0)$
For $y$-intercept $x=0$
$y=-(0+1)^2$
$y=-1$
$y$-intercept is $A(0,-1)$
In this case we have quadratic function which has a form of parabola. The $-$ sign in front of the equation informs us that the parabola is concave down. And as we have only one $x$-intercept, we can assume that the point is vertex of parabola.
Let's find one more point for easier sketching
$x=-3$
$y=-(-3+1)^2$
$y=-4$
$C(-3,-4)$
We can now plot the points and approximately connect them, giving a form of parabola. See the image below.