# How do you find the domain and range of the function y = −2x^2 − 6x + 1 without graphing?

Mar 26, 2015

For the domain, I ask myself what real numbers I could use for $x$ and get a real number for $y$. There's no division by $x$ , so no danger of dividing by $0$. And there are no even roots, so I there's no chance of getting an imaginary answer.

The domain is all real numbers. $\left(- \infty . \infty\right)$

To find the range (without thinking about the graph) I need to know what real numbers could I use for $y$ and get a real number for $x$

To do that, I'll sole the equation for $x$ is terms of $y$.

Solve for $x$:
$y = - 2 {x}^{2} - 6 x + 1$

$2 {x}^{2} + 6 x + y - 1 = 0$

Use the quadratic formula with $a = 2$, $b = 6$, $c = y - 1$

$x = \frac{- 6 \pm \sqrt{{\left(6\right)}^{2} - 4 \left(2\right) \left(y - 1\right)}}{4}$

This will give real numbers for $x$ if $36 - 8 y + 8 \ge 0$

$44 \ge 8 y$

$y \le \frac{11}{2}$

The range is all reals less than or equal to $\frac{11}{2}$.
The interval $\left(- \infty , \frac{5}{2}\right)$

Mar 26, 2015

$y = - 2 {x}^{2} - 6 x + 1$

Complete the square:

$y = - 2 \left({x}^{2} + 3 x\right) + 1$

(If I knew how, I'd leave some space between "$3 x$" and the "$\text{)}$" )

$y = - 2 \left({x}^{2} + 3 x + \left(\frac{9}{4}\right) - \left(\frac{9}{4}\right)\right) + 1$

$y = - 2 \left({x}^{2} + 3 x + \frac{9}{4}\right) + \frac{9}{2} + 1$

$y = - 2 {\left(x + \frac{3}{2}\right)}^{2} + \frac{11}{2}$

Now for every real value for $x$, we get a real number for $y$, so the domain is $\left(- \infty , \infty\right)$ (All real numbers.)

Furthermore, for $x = - \frac{3}{2}$, the first term on the right, $- 2 {\left(x + \frac{3}{2}\right)}^{2}$ is $0$, so we get $y = \frac{11}{2}$.

Finally, for all real $x$, the term $- 2 {\left(x + \frac{3}{2}\right)}^{2} \le 0$ so we will be getting $y$ values less than or equal to $\frac{11}{2}$.

The range is $\left(- \infty , \frac{11}{2}\right]$