Question 1161846
The solution is the area between the parabola {{{y=x^2-8}}} and the straight line {{{x+y=1}}} , including points on the line, but not including points on the parabola.
{{{drawing(300,300,-10,10,-10,10,
grid(0),
blue(line(-10,11,10,-9)),
blue(line(-4.5,12.25,-4,8)),
blue(line(-3.5,4.25,-3,1)),
blue(line(-2.5,-1.75,-2,-4)),
blue(line(-1.5,-5.75,-1,-7)),
blue(line(-0.5,-7.75,0,-8)),
blue(line(0.5,-7.75,1,-7)),
blue(line(1.5,-5.75,2,-4)),
blue(line(2.5,-1.75,3,1)),
blue(line(3.5,4.25,4,8)),
green(line(-3.5,4.5,-3.5,4.25)),
green(line(-3,1,-3,4)),
green(line(-2.5,3.5,-2.5,-1.75)),
green(line(-2,-4,-2,3)),
green(line(-1.5,2.5,-1.5,-5.75)),
green(line(-1,2,-1,-7)),
green(line(-0.5,1.5,-0.5,-7.75)),
green(line(0,-8,0,1)),
green(line(0.5,0.5,0.5,-7.75)),
green(line(1,-7,1,0)),
green(line(1.5,-0.5,1.5,-5.75)),
green(line(2,-4,2,-1)),
green(line(2.5,-1.5,2.5,-1.75)),
locate(4,8,y=x^2-8),
locate(4.8,-7,x+y=1)
)}}}
The boundary lines intersect at the solution of {{{system(y=1-x,y=x^2*8)}}} ,
where {{{x^2-8=1-x}}} -->  {{{x*2+x-9=0}}} --> {{{highlight(x=(-1 +- sqrt(37))/2)}}}
The approximate coordinate values for those points are
{{{x=2.541}}} with {{{y=2.541^2-8=-1.543}}}
and
{{{y=-3.541^2-8=-1.543}}} with {{{y=(-32.541)^2-8=4.539}}}