The intercepts cross an axis. 

For the $\displaystyle y$ intercept, set $\displaystyle x=0$ to get $\displaystyle y=6$

For the $\displaystyle x$ intercept, set $\displaystyle y=0$ to get $\displaystyle x=8$

So the sum of the intercepts is $\displaystyle 14$.

To find the intercepts of a line, we must set the $\displaystyle x$ and $\displaystyle y$ values equal to zero and then solve.  

$\displaystyle 0 = 3x +6$

$\displaystyle -6 = 3x$

$\displaystyle x = -2$

$\displaystyle (-2, 0)$

$\displaystyle y = 3 (0) + 6$

$\displaystyle y = 6$

$\displaystyle (0, 6)$

The $\displaystyle x$-intercept is the point where the y-value is equal to 0. Therefore,

$\displaystyle 0=4x-8$

$\displaystyle 4x=8$

$\displaystyle x=2$

To find the x-intercept, we simply plug $\displaystyle y=0$ into our function. giving us $\displaystyle 0=x^2-6x+9$. We can factor that equation, making it $\displaystyle 0=(x-3)^2$. We can not solve for $\displaystyle x$, and we get $\displaystyle x=3$. To find the y-intercept, we do the same thing, however this time, we plug in $\displaystyle x=0$ instead. This leaves us with $\displaystyle y=9$. With an x-intercept of $\displaystyle 3$ and a y-intercept of $\displaystyle 9$, it is clear that the y-intercept is greater than the x-intercept.

To find the x-intercept, set y equal to 0.

$\displaystyle 0 = 3x-9$

Now solve for x by dividing by 3 on both sides.

$\displaystyle 9 = 3x$

$\displaystyle \frac{9}{3}=\frac{3}{3}x$

This reduces to,

$\displaystyle 3=x$

To find the y-intercept, set x equal to 0.

$\displaystyle y = 0^2 +0 - 9$

Now solve for y.

$\displaystyle y=-9$

The x-intercept of a line is the x-value where the line hits the x-axis. This occurs when y is 0. To determine the x-value, plug in 0 for y in the original equation, then solve for x:

$\displaystyle \small 0 = 2x - 5$ add 5 to both sides

$\displaystyle \small 5 = 2x$ divide by 2

$\displaystyle \small 2.5 = x$

The x-intercepts of any curve are the x-values where the curve is intersecting the x-axis. This happens when y = 0. To figure out these x-values, plug in 0 for y in the original equation and solve for x:

$\displaystyle \small (x-4)^2 + 0^2 = 9$ adding 0 or 0 square doesn't change the value

$\displaystyle \small (x-4)^2 = 9$ take the square root of both sides

$\displaystyle \small x - 4 = \pm 3$ this means there are two different potential values for x, and we will have to solve for both. First:

$\displaystyle x - 4 = -3$ add 4 to both sides

$\displaystyle \small x = 1$

Second: $\displaystyle \small x - 4 = 3$ again, add 4 to both sides

$\displaystyle x = 7$

Our two answers are $\displaystyle x = 1$ and $\displaystyle x = 7$.

To find where the graph hits the y-axis, plug in 0 for x:

$\displaystyle \small (y+3)^2 - (0-2)^2 = 21$ first evaluate 0 - 2 

$\displaystyle \small (y+3)^2 - (-2)^2 = 21$ then square -2

$\displaystyle \small (y+3)^2 - 4 = 21$ add 4 to both sides 

$\displaystyle \small (y+3)^2 = 25$ take the square root of both sides

$\displaystyle \small y + 3 = \pm 5$ now we have 2 potential solutions and need to solve for both

a) $\displaystyle \small y + 3 = 5$

$\displaystyle \small y = 2$

b) $\displaystyle \small y +3 = -5$

$\displaystyle \small y = -8$

The y-intercept(s) occur where the graph intersects with the y-axis. This is where x=0, so we can find these y-values by plugging in 0 for x in the equation:

$\displaystyle \small y = 0^2 - 16$

$\displaystyle \small y = -16$

The x-intercept(s) occur where the graph intersects with the x-axis. This is where y=0, so we can find these x-values by plugging in 0 for y in the equation:

$\displaystyle \small 0 = x^2 - 16$ add 16 to both sides

$\displaystyle \small 16 = x^2$ take the square root

$\displaystyle \small \pm 4 = x$