Question 701011
1.

{{{ y=x^2-5x-6}}}

the intercepts:

set {{{x=0}}} and find {{{y-intercept}}}

{{{ y=0^2-5*0-6}}}

{{{ highlight(y=-6)}}}.......{{{y-intercept}}} is at ({{{0}}},{{{-6}}})


set {{{y=0}}} and find {{{x-intercept}}}

{{{ 0=x^2-5x-6}}}.....use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ...in your case {{{a=1}}}, {{{b=-5}}}, and {{{c=-6}}}

{{{x = (-(-5) +- sqrt( (-5)^2-4*1*(-6) ))/(2*1) }}} 

{{{x = (5 +- sqrt( 25+24 ))/2 }}} 

{{{x = (5 +- sqrt( 49 ))/2 }}}

{{{x = (5 +- 7)/2 }}}

solutions:

{{{x = (5 +7)/2 }}}

{{{x = 12/2 }}}

{{{highlight(x = 6) }}}.......one {{{x-intercept}}}

and

{{{x = (5-7)/2 }}}

{{{x = -2/2 }}}

{{{highlight(x =-1) }}}.......another {{{x-intercept}}}

so, {{{x-intercepts}}} are at ({{{6}}},{{{0}}}) and ({{{-1}}},{{{0}}})

graph:

 {{{ graph(600, 600, -10, 10, -15, 10, x^2-5x-6) }}} 


parabola has an axis of symmetry which is the line that runs down its 'center' or vertex

since your equation is in standard form, then the formula for the axis of symmetry is:

    {{{x = -b/2a}}} from the general standard form equation {{{y = ax^2+bx + c}}}

so, plug in {{{a=1}}} and {{{b=-5}}}

{{{x = -(-5)/2*1}}}

{{{x = 5/2}}}

{{{x = 2.5}}}....so,axis of symmetry is a line parallel to {{{y-axis}}} and crosses {{{x-axis}}} at point ({{{2.5}}},{{{0}}})

let's see it on a graph:



{{{drawing(600, 600, -10, 10, -16, 10,  blue(line(2.5,10,2.5,-16)), grid(0),
graph( 600, 600, -10, 10, -16, 10,x^2-5x-6)) }}}


2.

{{{x^2+y^2+14x-12y=-76}}}

{{{x^2+y^2+14x-12y+76=0}}}......write {{{76}}} as {{{49+36-9}}}

{{{x^2+14x+y^2-12y+49+36-9=0}}}...group {{{49}}} with {{{x^2+14x}}}, and {{{36}}} with {{{y^2-12y}}}

{{{(x^2+14x+49)+(y^2-12y+36)-9=0}}}...recognize square of the sum for {{{x}}} part and square of the difference  for {{{y}}} part

{{{(x+7)^2+(y-6)^2-9=0}}}.....move {{{-9}}} to the right

{{{(x+7)^2+(y-6)^2=9}}}.........your answer for the equation of the circle in standard form  {{{ (x-h)^2+(y-k)^2=r^2}}}

as you can see {{{h=-7}}} and {{{k=6}}}, so the center is at ({{{-7}}},{{{6}}})

 radius is {{{r=3}}}

 intercepts:


set {{{y=0}}} and find {{{x-intercepts}}}

{{{(x+7)^2+(0-6)^2=9}}}

{{{(x^2+14x+49)+(-6)^2=9}}}

{{{x^2+14x+49+36=9}}}

{{{x^2+14x+85-9=0}}}

{{{x^2+14x+76=0}}}.......use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-14 +- sqrt( 14^2-4*1*76 ))/(2*1) }}}

{{{x = (-14 +- sqrt( 196-304 ))/2 }}}

{{{x = (-14 +- sqrt(-108 ))/2 }}}

{{{x = (-14 +- 10.39*i)/2 }}}....as you can see, solution is complex, and it means there are {{{no}}} {{{x-intercepts}}}


set {{{x=0}}} and find {{{y-intercepts}}}

{{{(0+7)^2+(y-6)^2=9}}}

{{{49+(y^2-12y+36)-9=0}}}

{{{y^2-12y+36+49-9=0}}}

{{{y^2-12y+76=0}}}.......use quadratic formula

{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{y = (-12 +- sqrt( 12^2-4*1*76 ))/(2*1) }}}

{{{y = (-12 +- sqrt( 144-304 ))/2 }}}

{{{y = (-12 +- sqrt(-108 ))/2 }}}

{{{x = (-12 +- 10.39*i)/2 }}}...solution is complex, and it means there are {{{no}}} {{{y-intercepts}}}


and graph of the circle:

{{{drawing( 600, 600, -15, 5, -5, 10,
  grid( 1 ),circle(-7,6,0.1),locate(-7,6,O(-7,6)),
  circle( -7, 6, 3 )
 
))}}}