Question 1113225
{{{x^2-3xy+y^2+10x-10y+21=0}}}
<pre>
We solve for y:

{{{y^2-3xy-10y+x^2+10x+21=0}}}

{{{y^2+(-3xy-10y)+(x^2+10x+21)=0}}}

{{{y^2+(-3x-10)y+(x^2+10x+21)=0}}}

a=1, b=(-3x-10), c=(x²+10x+21)

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

{{{y = (-(-3x-10) +- sqrt( (-3x-10)^2-4(1)(x^2+10x+21)))/(2(1)) }}}

{{{y = (3x+10 +- sqrt( (-3x-10)(-3x-10)-4(x^2+10x+21)))/2 }}}

{{{y = (3x+10 +- sqrt( 9x^2+60x+100-4x^2-40x-84))/2 }}}

{{{y = (3x+10 +- sqrt( 5x^2+20x+16))/2 }}}

Make a table of values using that formula

 x |      y
-10|-1.1 and -18.9
 -7|  0  and -11 
 -4|  1  and  -3
 -3|  1  and   0
 -2|imaginary
 -1|  4  and   3
  0|  7  and   3
  3| 15  and   4
 11| 36  and   7

Plotting some of those and others gives the rotated
and shifted hyperbola:

{{{drawing(400,200,-10,10,-5,5,

graph(400,200,-10,10,-5,5,(1/2)(sqrt(5x^2+20x+16)+3x+10)),
graph(400,200,-10,10,-5,5,(-1/2)(sqrt(5x^2+20x+16))+3x/2+5)) )}}}

Edwin</pre>