# How do you find the domain and range of f(x)=5x^2+2x-1?

Aug 8, 2015

The domain is the set of all real numbers $\mathbb{R} = \left(- \infty , \infty\right)$ and the range is the set of all real numbers greater than or equal to $- \frac{6}{5}$: $\left[- \frac{6}{5} , \infty\right)$.

#### Explanation:

The function $f \left(x\right) = 5 {x}^{2} + 2 x - 1$ is quadratic and any number can be plugged into it. Therefore, its (natural) domain is $\mathbb{R} = \left(- \infty , \infty\right)$.

The graph of $f$ is a parabola opening upward. To find the range, we must find the lowest point. The quickest way to find this is with calculus, but since this is an algebra question we will use the method of "completing the square" instead.

First, factor the coefficient $5$ of ${x}^{2}$ out of the first two terms like this:

$f \left(x\right) = 5 {x}^{2} + 2 x - 1 = 5 \left({x}^{2} + \frac{2}{5} x\right) - 1$.

Next, take the coefficient of $x$, $\frac{2}{5}$, divide it by 2 to get $\frac{1}{5}$, then square that number to get $\frac{1}{25}$. Put this number inside the parentheses above and compensate for this by subtracting $5 \cdot \frac{1}{25} = \frac{1}{5}$ outside the parentheses:

$f \left(x\right) = 5 \left({x}^{2} + \frac{2}{5} x\right) - 1 = 5 \left({x}^{2} + \frac{2}{5} x + \frac{1}{25}\right) - 1 - \frac{1}{5}$

The reason this trick is a good idea is the resulting expression inside the parentheses is now a perfect square. Factoring it gives:

$f \left(x\right) = 5 {\left(x + \frac{1}{5}\right)}^{2} - \frac{6}{5}$.

Once the function is in this form ("vertex form"), it is clear that the lowest possible output for $f$ is $- \frac{6}{5}$, which occurs when $x = - \frac{1}{5}$. In other words, the (low) vertex of the parabola is the point with coordinates $\left(x , y\right) = \left(- \frac{1}{5} , - \frac{6}{5}\right)$.

Since the function $f$ also grows without bound as $x \to \pm \infty$, it follows that the range (the set of all possible outputs as $x$ varies over the domain) is the interval $\left[- \frac{6}{5} , \infty\right)$.

Graph the function on your calculator to confirm this.