Question 136598
Ok, if you are confused by graphing, you need to go back to the basics.


Anything that looks like this: (x,y) is called an ordered pair.  It is called 'ordered' because the x value always comes first and the y value second.  So if you see (-1,3), you know that {{{x=-1}}} and {{{y=3}}} because of the order that the numbers are specified.  Every ordered pair corresponds to a point on the xy plane, also called {{{R^2}}}.  It is called {{{R^2}}} because each of the axes is just a copy of the real number line.  (As you might suspect, the number line is just {{{R^1}}}, and three-dimensional space is {{{R^3}}}).  For our purposes here, all you need to remember is that when you want to find a point on the graph given an ordered pair, you need to take the x-coordinate, that is the first number in the pair, and go either left (-) or right (+) from the origin.  And then you take the y-coordinate, the second number in the pair, and either go down (-) or up (+).  For our example, (-1,3), you would go left 1 unit and up 3 units.


An equation in x and y, such as {{{y=2x+4}}} or {{{y=x^2-4x+4}}} is just a rule that defines a set of ordered pairs that creates a line in {{{R^2}}} (a straight line in the first case, a parabola in the second).  And the relationship between the graph and the equation is this:  If some ordered pair is on the graph, then the coordinates of that ordered pair make the equation a true statement, and vice versa.  If some ordered pair is NOT on the graph, then the coordinates of the ordered pair will make the equation a false statement.


Let's look at your equation:  {{{x^2+x=12}}}.


First of all, you don't have a y variable, so now what?  If you add -12 to both sides of your equation, then you have {{{x^2+x-12=0}}}.  Notice that this is just the same as saying {{{y=x^2+x-12}}} AND {{{y=0}}}.   Now you have an equation in x and y that can be graphed.


Any equation in the form {{{y=ax^2+bx+c}}} with a, b, and c as real coefficients graphs as a parabola -- a big U-shaped thing that opens upward if a is positive and downward if a is negative.  In the case of your equation, {{{a=1}}}, {{{b=1}}}, and {{{c=-12}}}.


The trick to graphing a parabola is to find the vertex, that is the point at the very bottom of the valley (or the very top of the hill if you have a downward opening graph).  This is actually quite simple:  The x-coordinate of the vertex is given by {{{-b/2a}}}.  Plug in your values for a and b, you get {{{-1/2}}}.  The y-coordinate of the vertex is found by substituting the x-coordinate into the equation and doing the arithmetic.  {{{y=(-1/2)^2+(-1/2)-12=1/4-1/2-12=-49/4}}}  Now we know the ordered pair that represents the vertex: ({{{-1/2)}}},{{{-49/4}}}), and you can plot that on your graph.


{{{drawing(800,800,-15,15,-15,15,
grid(1),
green(circle(-1/2,-49/4,.2)),
locate(.1,-25/2,V(-1/2,-49/4))
)}}}


Now let's move one and one half units to the right from the x-coordinate of the vertex.  That takes us to {{{x=1}}}.  Substitute 1 for x in the equation:  {{{y=(1)^2+(1)-12=1+1-12=-10}}}.  That gives us a second point on the graph, namely, (1,-10).  And actually, that gives us a third point right away because a parabola has symmetry around the vertical line that passes through the vertex.  That means that if we know the value of y for an x value some distance to the right of the vertex, that same value of y will be true for an x value that same distance to the left of the vertex.   So, one and one-half units to the left of {{{-1/2}}} is {{{-2}}}, so our third point is ({{{-2}}},{{{-10}}}).  Let's plot those two points:



{{{drawing(800,800,-15,15,-15,15,
grid(1),
green(circle(-1/2,-49/4,.2)),
locate(.1,-25/2,V(-1/2,-49/4)),
green(circle(1,-10,.2)),
locate(1.6,-10.25,P1(1,-10)),
green(circle(-2,-10,.2)),
locate(-6,-10.25,P2(-2,-10))
)}}}


Let's move one more unit left and right.  I'll let you do the arithmetic, but here is the picture.


{{{drawing(800,800,-15,15,-15,15,
grid(1),
green(circle(-1/2,-49/4,.2)),
locate(.1,-25/2,V(-1/2,-49/4)),
green(circle(1,-10,.2)),
locate(1.6,-10.25,P1(1,-10)),
green(circle(-2,-10,.2)),
locate(-6,-10.25,P2(-2,-10)),
green(circle(2,-6,.2)),
locate(2.6,-6.25,P3(2,-6)),
green(circle(-3,-6,.2)),
locate(-7,-6.25,P2(-3,-6))
)}}}


One more time, but let's move two units to the right, from 2 to 4:


{{{drawing(800,800,-15,15,-15,15,
grid(1),
green(circle(-1/2,-49/4,.2)),
locate(.1,-25/2,V(-1/2,-49/4)),
green(circle(1,-10,.2)),
locate(1.6,-10.25,P1(1,-10)),
green(circle(-2,-10,.2)),
locate(-6,-10.25,P2(-2,-10)),
green(circle(2,-6,.2)),
locate(2.6,-6.25,P3(2,-6)),
green(circle(-3,-6,.2)),
locate(-7,-6.25,P4(-3,-6)),
green(circle(4,8,.2)),
locate(4.6,7.75,P5(4,8)),
green(circle(-5,8,.2)),
locate(-9,7.75,P6(-5,8))
)}}}


That's enough points to get a good sketch.  Let's draw a smooth curve through the points:


{{{drawing(800,800,-15,15,-15,15,
grid(1),
green(circle(-1/2,-49/4,.2)),
locate(.1,-25/2,V(-1/2,-49/4)),
green(circle(1,-10,.2)),
locate(1.6,-10.25,P1(1,-10)),
green(circle(-2,-10,.2)),
locate(-6,-10.25,P2(-2,-10)),
green(circle(2,-6,.2)),
locate(2.6,-6.25,P3(2,-6)),
green(circle(-3,-6,.2)),
locate(-7,-6.25,P4(-3,-6)),
green(circle(4,8,.2)),
locate(4.6,7.75,P5(4,8)),
green(circle(-5,8,.2)),
locate(-9,7.75,P6(-5,8)),
graph(800,800,-15,15,-15,15,x^2+x-12)
)}}}


Now, going back to your original problem, you wanted to solve {{{x^2+x=12}}}, which we re-wrote as {{{x^2+x-12=0}}}.  In other words, we want to find the values of x that will make y be 0.  Can you see them on the graph?  Remember, {{{y=0}}} for any point that is on the x-axis, so the values of x that make {{{y=0}}} are the x-coordinates of the points where the graph intersects the x-axis.



{{{drawing(800,800,-15,15,-15,15,
grid(1),
green(circle(-1/2,-49/4,.2)),
locate(.1,-25/2,V(-1/2,-49/4)),
green(circle(1,-10,.2)),
locate(1.6,-10.25,P1(1,-10)),
green(circle(-2,-10,.2)),
locate(-6,-10.25,P2(-2,-10)),
green(circle(2,-6,.2)),
locate(2.6,-6.25,P3(2,-6)),
green(circle(-3,-6,.2)),
locate(-7,-6.25,P4(-3,-6)),
green(circle(4,8,.2)),
locate(4.6,7.75,P5(4,8)),
green(circle(-5,8,.2)),
locate(-9,7.75,P6(-5,8)),
graph(800,800,-15,15,-15,15,x^2+x-12),
red(circle(3,0,.2)),
locate(3.6,1,Z1(3,0)),
red(circle(-4,0,.2)),
locate(-8,1,Z2(-4,0))
)}}}


There are a couple of ways to check your solutions algebraically.  The first and simplest is to take the two values you read off of the graph and substitute each those values into the original equation.  If you get a true statement for both answers, then you have proven that the values are solutions to the equation.


The second way is to solve the equation algebraically.  Starting with the equation in standard form:  {{{x^2+x-12=0}}}, you can either factor the trinomial and use the Zero Product Rule, complete the square, or use the quadratic formula.  Any of these methods will result in {{{x=-4}}} or {{{x=3}}}.