Question 701011
1. {{{y=x^2-5x-6}}} can be called a quadratic function and graphs as a parabola opening up. (It's shaped like a smile).
 
Intercepts are the points where the graph representing that equation crosses the x-axis and the y-axis.
The x-axis is the set of all the points with {{{y=0}}}.
Making {{{y=0}}} we find the x-intercepts:
{{{0=x^2-5x-6}}} --> {{{0=(x-6)(x+1)}}} (factoring)
has a solution when {{{x-6=0}}} --> {{{highlight (x=6)}}} and
a solution when {{{x+1=0}}} --> {{{highlight(x=-1)}}}
The y-axis is the set of all the points with {{{x=0}}} .
Making {{{x=0}}} we find the y-intercept:
{{{y=0^2-5*0-6}}} --> {{{y=0-1-6}}} --> {{{highlight(y=-6)}}}
 
When teachers say to test for SYMMETRY they usually want you to find out, without graphing,
if the graph representing the equation has symmetry.
Maybe the side to the right of the y-axis is a mirror image of the left side.
Or maybe the side above of the x-axis is a mirror image of side below the x-axis.
Or maybe every point has a reflection on the other side of the origin.
To find out, you change {{{x}}} to {{{-x}}},
or {{{y}}} to {{{-y}}},
or (x,y) to (-x,-y),
and see if you end up with the same equation.
For {{{y=x^2-5x-6}}}, any of those changes gives you a different equation.
The graph for {{{y=x^2-5x-6}}} has an axis of symmetry, but it is neither the x-axis, not the y-axis.
 
Knowing that a graph will be symmetrical before you start graphing, makes graphing easier.
In pictures,
{{{graph(100,100,-5,5,-5,5,x^2/3)}}} is a function with symmetry about the y-axis because it has the same y for {{{x}}} and {{{-x}}} ( f(x)=f(-x) }
{{{x^2-6x+y^2=0}}} graphs as {{{drawing(100,100,-2,8,-5,5,grid(1),blue(circle(3,0,3)))}}} and has symmetry about the x-axis, because for {{{y}}} and for {{{-y}}} the equation is the same.
{{{x^2+y^2=16}}} {{{drawing(100,100,-6,6,-6,6,grid(1),blue(circle(0,0,4)))}}} and {{{xy=1}}} {{{graph(100,100,-5,5,-5,5,1/x)}}}
have symmetry about the origin because the equation is the same for point (x,y) as for point (-x,-y).
 
2. {{{x^2+y^2+14x-12y=-76}}} --> {{{x^2+14x+y^2-12y=-76}}} --> {{{x^2+14x+49+y^2-12y+36=-76+49+36}}} --> {{{(x^2+14x+49)+(y^2-12y+36)=9}}} --> {{{highlight((x+7)^2+(y-6)^2=3^2)}}}
That equation is true for all the points (x,y) that are at distance 3 from point (-7,6).
That is the equation of a circle with radius 3, centered at the point (-7,6),
with {{{x=-7}}} and {{{y=6}}}.
The center is too far below the x-axis compared to the radius, so the circle does not cross the x-axis, and there are no x-intercepts.
The same thing happens with the y-axis. The center is too far to the right. There is no y-intercept.