How do I find the range of $y = x^2- x - 2$ without graphing? I can't use the quadratic formula or completing the square to find the range for y because y is not equal to zero and I am not allowed to solve this question using graphing. How do I solve it?
By the way, the range is [-9/4,+∞]. I'm just not sure how to get to it.
 A: Notice $x^2-x-2=(x-\frac12)^2-\frac94$. The squared term is greater than or equal to $0$ and indeed can take any nonnegative value, so the range is $[-9/4,\infty)$. 
A: $$x^2-x-2=(x-1/2)^2-9/4\ge -9/4$$
The range is $[-9/4,\infty)$
A: All quadratic functions have minimums m (or maximums) that happen at their vertex. These functions are symmetric around their vertex (draw a graph to see).
This means that the vertex is located at the mid point of the line segment joining the x-intercepts. So find zeros:
$y=(x-2)(x+1)$ and the zeros are $2$ and $-1$. So $2+(-1)$ divided by $2$ is $0.5$. So the minimum (since the parabola opens upward because of the positive leading coefficient) is at $(0.5,y)$
So your range is anything greater than that $y$ coordinate.
A: Since you are asking on how to get to $(x-\frac12)^2-\frac94$.
This method is called 'completing the square' and is taught in school.
The basic idea is to use a binomial formula to get an expression like $a(x+b)^2+c$ from a quadratic polynomial.
This is used to find the vertex, which is then at the point $(-a|c)$.
If the sign of the x^2 is positiv, this is a minimum. If the sign is negativ it is a maximum, and like this we can figure out the range. (If the domain is $\mathbb{R}$)
How do we do this?
We have to follow the following steps:


*

*The coefficient of $x^2$ has to be $1$

*Take the coefficient of $x$. If it is positiv we apply the first binomial formula [$(a+b)^2=a^2+2ab+b^2$], if it is negativ the second binomial formula [$(a-b)^2=a^2-2ab+b^2$]. For that half the coefficient, square it, add and subtract the result.

*Evaluate a binomial formula.


So take $x^2-x-2$. Obviously the coefficent of $x^2$ is already $1$. So there is nothing to do in step 1. Else we would have to factor this out.
Step 2:
$-x$ has coefficient $(-1)$, so we will apply the 2nd binomial formula.
As I said we have to add and subtract the square of the half of (-1). That is $(\frac{-1}2)^2=\color{red}{\left(\frac12\right)^2}=\frac14$.
We get:
$x^2-x+\underbrace{\frac14-\frac14}_{=0}-2$

Notice that it is important to add and subtract. Basically we add a $+0$, which is elegantly written, so we can apply the binomial formula. 

If we would just add something and not subtract, then we would change the term and get a wrong result. 
Now we apply the binomial formula. 
$\underbrace{x^2-x+\color{red}{\left(\frac12\right)^2}}_{\text{2nd binomial formula}}-\frac14-2=(x-\color{red}{\frac12})^2-\frac94$

Notice that the $\color{red}{1/2}$ is constructed such that the binomial is completed with exactly this result (see step 2), which makes it easy to complete the square.

Now we have $(x-\frac12)^2-\frac94\geq -\frac94$, because $(x-\frac12)^2$ is always positive (or to say it better nowhere negativ, because the expression can take the value $0$), so we can not get smaller then $-\frac94$, but arbitraly big. 
So the range in question is indeed $[-\frac94,\infty)$, where we take the minimal value $-\frac94$ at the vertex, that we can now see from the formula above:
$(x-\frac12)^2-\frac94\leadsto\text{Vertex at $(\frac12|-\frac94)$}$
A: Let $f(x)=x^2-x-2$.
When $y=f(x)$ is in the range of $f$ then the quadratics $x^2-x-2-y=0$ has real solutions, so the discriminant is positive, and vice-versa, for any positive discriminant, we get some $y$ in the range of $f$.
$\Delta=1^2-4(-2-y)=9+4y\ge 0\iff y\ge -\dfrac 94$
So the range is $[-\frac 94,+\infty)$
