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

\(\displaystyle \small \pm 4 = x\)