Answer: The points of intersection with the axes are \((-1,0)\), \((2,0)\) and \((0,-1)\).
The curve satisfies \(-1 \leqslant y \leqslant 15\) for all real \(x\).
The turning points are \((0,-1)\) and \((-4,15)\).
The horizontal asymptote is \(y=5\).
We start with \(y=\frac{5(x^2-x-2)}{x^2+5x+10}\).
Intersections with the axes
For the x-axis, \(y=0\), so the numerator must be zero:
\(5(x^2-x-2)=0 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0\).
Hence \(x=2\) or \(x=-1\), giving the points \((2,0)\) and \((-1,0)\).
For the y-axis, \(x=0\):
\(y=\frac{5(0-0-2)}{0+0+10}=\frac{-10}{10}=-1\).
So the y-intercept is \((0,-1)\).
Showing that \(-1 \leqslant y \leqslant 15\)
Rearrange the equation to make a quadratic in \(x\):
\(y(x^2+5x+10)=5(x^2-x-2)\).
So
\(yx^2+5xy+10y=5x^2-5x-10\),
and therefore
\((y-5)x^2+(5y+5)x+10(y+1)=0\).
For real values of \(x\), the discriminant must be non-negative:
\((5y+5)^2-4(y-5)\cdot 10(y+1) \geqslant 0\).
Simplifying:
\(25(y+1)^2-40(y-5)(y+1)\geqslant 0\).
Factor out \(5(y+1)\):
\(5(y+1)[5(y+1)-8(y-5)]\geqslant 0\).
So
\(5(y+1)(45-3y)\geqslant 0\),
which gives
\((y+1)(15-y)\geqslant 0\).
Hence
\(-1 \leqslant y \leqslant 15\).
Turning points
The endpoints of the range occur at turning points. Since \(y=-1\) gives \(x=0\), one turning point is \((0,-1)\).
Since \(y=15\), substitute into the rearranged quadratic:
\((15-5)x^2+(5\cdot 15+5)x+10(15+1)=0\)
\(10x^2+80x+160=0\)
\(x^2+8x+16=0\)
\((x+4)^2=0\), so \(x=-4\).
Thus the other turning point is \((-4,15)\).
Horizontal asymptote
The highest powers of \(x\) in numerator and denominator are both \(x^2\), so the horizontal asymptote is the ratio of leading coefficients:
\(y=\frac{5}{1}=5\).
Sketch
The curve crosses the x-axis at \((-1,0)\) and \((2,0)\), crosses the y-axis at \((0,-1)\), has turning points at \((-4,15)\) and \((0,-1)\), and approaches the horizontal asymptote \(y=5\).
