How many times will they intersect?

We know that $| \sin x | \leq 1$ , so we need only consider the portion of the graph of $y=x^{2}+x+1$ where $y \leq 1$ .

We also know that the parabola formed by $y=x^{2}+x+1$ must open upwards and have a relative (as if we consider $y=ax^{2}+bx+c$ , $a=1$ .)

Now we can find where $x^{2}+x+1 \leq 1$ by setting:

$x^{2}+x+1=1$
$x^{2}+x=0$
$(x)(x+1)=0$
$x=0$ or $x=-1$

Also, for values $-1<x<0$ , we know that $x^{2}+x+1>0$ , as $x^{2}>0$ and $x+1>0$ .

Now to consider $y= \sin x$ . For the interval we're looking at ( $-1 \leq x \leq 0$ ), we observe that $-\frac{\pi}{2}<x \leq 0$ , and for those values, $\sin x \leq 0$ .

Over the only values where $y=\sin x$ and $y=x^{2}+x+1$ could possibly intersect, they do not, as $y=\sin x$ is negative or zero, and $y=x^{2}+x+1$ is positive.

we can write , x^2+x+1=(x+1/2)^2 +3/4; so x^2 +x+1 is always positive. it's minimum value is 3/4 which occurs at x=(-1/2), also at x=0 given expression is equal to 1. we know that sin x always lie between [-1 ,1]. for the intersection of sin x and x^2 +x+1 we need to consider only +ve part of sin x ,since given quadratic expression is always +ve. again note that for (x>0) given expression is always greater 1,hence they will never intersect each other for +ve values of x. now consider x<0, at x=(-1/2) given expression =3/4 (minimum value) and between x=(-pi) to x=0 ,i.e between x=(-3.14) to x=0, sin x is less than zero, so they will not intersect, for any other values of x ,x^2 +x+1 is always greater than sin x. hence they will never intersect.