Sketching the graph from function I am having some issues with sketching the following graph.  Any assistance is appreciated.
The function $g(x) = -x^2 + 6x + 4$ , $K \leq x \leq 6$ where $K$ is a constant.
Also, $g^{-1}(x)= -(y+3)^2 + 13$.
How do I sketch the graph of $y=g(x)$ and $y=g^{-1}(x)$?

I have already completed the square for both, but I'm stuck on what to do after?
 A: Some Hints:
First, as pointed by the comments, the inverse function in the question is incorrect.
You need to find the correct inverse.
Find the inverse function $f^{-1}(x)$ by solving the following for $x$:
$$-x^2+6x+4=y$$
The above can be written as:
$$-x^2+6x+(4-y)=0$$
Solve the above using Quadratic formula and substitute $y=x$, to get two functions:
$$f_1^{-1}(x)=3+\sqrt{13+x}$$
and
$$f_2^{-1}(x)=3-\sqrt{13+x}$$
The above was to fix the inverse function. Now we look into how to draw the original function.
Find $f(0)$ to find the intercept of $f(x)$ with the y-axis to get the point: $(0,4)$
Solve $f(x)=0$ to find the intercepts of $f(x)$ with the x-axis using the quadratic formula:
$$x=\frac{-6+\sqrt{36+16}}{-2}$$
and 
$$x=\frac{-6-\sqrt{36+16}}{-2}$$
Now, we can find the function extreme points (maximum and/or minimum):
Find f'(x):
$$f'(x)=-2x+6$$
$f'(x)=0$ at extreme points, solving:
$$f'(x)=-2x+6=0$$
You get the point $(3,13)$ as an extreme point.
Since $f''(3)=-2*3=-6$ is less than zero, the point $(3,13)$ is at least a local maximum.
The last step is to generate a table of (x, f(x)) pairs before and after the roots and the maximum points then connect the values in the Parabola shape.
Since f(x) is a parabola, you can try to draw a parabola passing through the points identified above.
The inverse function(s) would be symmetrical across the line $y=x$.
I am not sure you can draw the inverse functions just from the above fact though.
Also, I did not see the value of the information: $K<=x<=6$. 

A: There's a bunch of online tools that can sketch these graphs for you.
The first graph is here.
The second graph is here.
How can you sketch these by hand yourself? For these simple curves you can:
1) Identify the curve if possible. In this case the highest order term is to the power of $2$, so we know that it's going to be a quadratic.
2) Find the intersection points with the axes. We can do this by setting $x=0$ and $y=0$ in turn.
3) For a more accurate sketch, you may wish to find local extrema by computing the derivatives. If you have already identified the curve, then you may only need to compute the first derivative.
This is definitely not an exhaustive list, and there are better approaches depending on the situation.
You may find this pdf on curve sketching helpful.
