Mathematics - Quadratic Equation Question with Solution | TestHub

Let $y=\frac{x^2+2x+1}{x^2+2x+7}$. We can rewrite this as $y=\frac{(x+1{)}^2}{(x+1{)}^2+6}$. Let $t=(x+1{)}^2$. Since $x$ is real, $t\ge 0$. So $y=\frac{t}{t+6}$. If $t=0$, $y=0$. As $t\to \infty$, $y=\frac{t}{t+6}=\frac{1}{1+6/t}\to 1$. Since $f(t)=\frac{t}{t+6}$ is an increasing function for $t\ge 0$, the range of $y$ is $[0,1)$.