I have 2 equations programmed, and want to find the points of intersection. I can see them, but not sure how to find the exact values of intersection. Can you help?

Since the y-variable is isolated in both equations, set the two right sides of the equations equal
x^2-7x+4=-8x^2+56x-86
Combine all like terms on one side of the equal sign
9x^2-63x+90=0. factor out the 9 to get 9(x^2-7x+10)=0
Now solve the equation (x^2-7x+10)=0. The solutions are x=5 and x=2
Calculate the y-value tof be y=-6
The two solutions are (2,-6) and (5,-6)
Graphically, here is how the two curves intersect. One intersection point in the screen capture is (2,-6). You can read that the other intersection point is (5,-6)

In theory yes. If you have a function y=f(x) and another y=g(x), an intersection point of the graphs of the two functions, if it exists, is a point (x_i,y_i) such that f(x_i)=g(x_i).
Graphically this means that (x_i, f(x_i))=(x_i, g(x_i)): The two curves pass through the same point in the Cartesian plane.
Now consider f(x_i)=g(x_i). That is an equation in x_i, or lest us just drop the _i to shorten and write f(x)=g(x), or f(x)-g(x)=0.
If you can feed it to the SOLVER in EQN, the equation mode, and the calculator gives you the roots of the equation, then for each root x_i found you have an intersection point (x_i, f(x_i)), or (x_i, g(x_i)). You can use either function to calculate the y-value since the two functions are supposed to be equal at the root x_i.
However it is much simpler to do with a graphing calculator.

In theory yes. If you have a function y=f(x) and another y=g(x), an intersection point of the graphs of the two functions, if it exists, is a point (x_i,y_i) such that f(x_i)=g(x_i).
Graphically this means that (x_i, f(x_i))=(x_i, g(x_i)): The two curves pass through the same point in the Cartesian plane.
Now consider f(x_i)=g(x_i). That is an equation in x_i, or lest us just drop the _i to shorten and write f(x)=g(x), or f(x)-g(x)=0.
If you can feed it to the SOLVER in EQN, the equation mode, and the calculator gives you the roots of the equation, then for each root x_i found you have an intersection point (x_i, f(x_i)), or (x_i, g(x_i)). You can use either function to calculate the y-value since the two functions are supposed to be equal at the root x_i.
However it is much simpler to do with a graphing calculator.

1. No solutions: The system is incoherent, incompatible Example: 2x+3y=8 and 2x+3y= 15. The two lines are parallel and distinct.
2. One solution: There exits a pair of values (x,y) that satisfy both linear equations. The two lines on a Cartesian graph have one intersection point.
3. Infinite number of solutions: The two equations are one and the same (one is just multiplied by some constant). The graph of the two lines yields the same line. One is superposed on the other. Any ordered pair (x,y) that satify one equation (there is an infinity of such pairs) satisfies the other.
4. Two solutions: cannot happen because the two lines can either intersect once, be parallel, or superposed one on the other.
   

I do not think you can directly since it is not a graphing calculator. However, if you are willing to put in the effort you can do it in two steps.
First write both equations in the form y=a_1x+b_1, and y2=a_2x+b_2
set the two right sides equal a_1x+b_1=a_2x+b_2. Isolate x by hand or use the solver of the calculator to find x. Substitute the value of x found in one of the equations y=ax+b. Calculate y. Your solution (the coordinates of the intersection point) is (x,y).

1. You draw two or more graphs.
2. After the graphs are displayed, press [2nd][TRACE] to access the (CALC)ulate menu.
   
3. Select [5:Interesct]
4. You will be prompted for a first curve: the equation of the curve will be displayed at the top left corner of the screen. If it is one the intersecting curves, press [ENTER]
5. You will be prompted for the second curve. (You can move from one curve to another by pressing the UpArrow or DownArrow).
6. After two curves are selected, you will be prompted for a guess for the X-value of an intersection point: you can use the keypad to enter a guess or use the left or right arrow to move the cursor towards a point of your choosing (if there are more than one point).
7. After a short while the calculator gives you a solution.
8. If it fails, you must make a better guess.
   

Half of the work involved in solving a problem is being able to formulate it so that it can be solved. I am afraid your formulation leaves too many details in the dark.
But I think I figured out what you want.
"You want to solve the equation graphically, ie find the zeros of a function"

FIRST METHOD

1. Create a single expression from the equation : gather all terms on one side so as to make "expression"=0
2. One such expression is X^2 +X -14=0
   
3. Draw the function y=X^2+X-14
4. Find the X-coordinates of the points where y=0

If that is the method you had in mind, it does not matter what specific values you set for Ymin and Ymax, because you are interested in Y values near Zero. You have to make sure the X-axis appears on the screen.
I assume you know how to use the [2nd][TRACE] (CALC) [2:Zero] function to find the zeros.

Here is the negative one Here is the positive one


SECOND METHOD
The second method entails

1. defining two functions, Y1=-X^2+5 and Y2=X-9,
2. Graphing the two functions.
3. Finding their intersections.
4. The X-Values of the two points of intersection of Y1 and Y2 are the solutions of the equation.

Here is the graph in standard window dimensions

As you see, one root does not show and you have either to Zoom out or move the Yrange downward (as seen on the right picture for which I set Ymin=-15, Ymax=5

I assume you know how to find the intersection of the two curves, and I will show you only one point.
To find the intersection you use the [2nd][TRACE] (CALC) [5:Intersect] command

Hi,
You should check your understanding of what a function is. You are drawing two functions the ranges of which do not overlap, since one branch is positive and the other is negative. You know that the only two points where there could be overlapping are the points where y=0 for both functions. Why would you need the calculator to confirm to you what you already know.
To define the two branches you had to take the square root of some expression say y= SQRT(5-x^2). That is a circle centered on O(0,0) with radius SQRT(5). The two points where the positive branch intersects the negative one are for y=0, meaning x1= SQRT(5) or x2= -SQRT(5).
What do you think is the exact value of SQRT(5): 2.236067977....? No, because SQRT(5) is an irrational number that has an infinite number of digits and no matter how many additional digits you may align to determine it will not make that representation the EXACT value of SQRT(5).

That does not mean you will never be able to find an intersection of the two curves. Maybe, if you take y=SQRT(4-x^ 2) the calculator will be able to find the intersections but that will remain one case.In general the calculator will not find the intersection.

I hope that I convinced that it is futile to seek, with the help of the calculator, the intersection of two irrational functions ( for they are irrational not rational as you claim) that share only two points.

Hope it helps.

Hello,
The function Intersect from the CALCULATE menu finf the coordinates of a point at which two or more curves intersect.
To use it:
1. Draw the functions.

2. Press [2nd][CALC][5:Intersect]



The cursor is on one of the curves. Read the equation top of the screen. If it is one of the curves you want press [ENTER]. The cursor jumps to another curve (in this case the only other curve).

Read the equation on top of the screen to verify thst it is the correct one. Press [ENTER]. The calculator asks asks for a guess of the coordinates of the intersection point.


As the intersection point is to the left of the current cursor position, use the left arrow to move cursor closer to the point.


Press [ENTER], and wait for the solution. Here it is.


In your question you talk about y intercept. If you want to calculate the ordinate of the point where a curve intersects the Y-axis, it is more efficient to use the [2nd][CAL][1:Value] selection.

You enter X=0 and press [ENTER]. The cursor jumps on the first curve (Y1=) an gives you the y-intercept.

Notice the position of cursor on graph. The y-value at the bottom is its ordinate.
To get the y-intercept of the second curve, leave the cursor on y axis and press the DownArrow. Cursor jumps to tthe second curve.
Since the X=0 is still stored, the value of y is displayed directly.



Hope it helps.

The calculator only displays from -10 to 10 on the x and y axes, and both of your lines are outside of that range. However, it is easy to change that.

1. Hit the gray "WINDOW" button that is directly under the screen
2. Change "Xmax" to 150
3. Change "Ymax" to 200

Remember to change both of those values back to 10 when you are done with this problem!

Here's how to graph it:
1. Hit the "Y=" button
2. Enter your first equation after Y1, and your second equation after Y2
3. Look over both the equations to make sure you typed them correctly
4. Hit the "GRAPH" button
5. If you don't see the intersection you will need to adjust Xmax and Ymax again
6. Hit Hit "2ND" and the "TRACE" to bring up the "CALC" menu
7. Scroll down to "intersect" and hit enter
8. Hit enter 3 times (You use the first and second curve options when you have more than 2 lines)
9. It should display the coordinates at the bottom
10. I got (76, 142.1) for the intersection

Thanks for asking this question! I used to do this a much harder way, but while I was calculating the answer to your question I hit a wrong button and accidentally found this method!