How do you find the domain and range of y = -2x^2 - 8 ?

Jun 14, 2017

domain$\to x \in \left(- \infty , + \infty\right)$
range $\to y \in \left(- \infty , + \infty\right)$

Explanation:

In the alphabet $\textcolor{red}{d}$ comes before $\textcolor{red}{r}$

So $\textcolor{red}{d}$omain comes before $\textcolor{red}{r}$ange

Input comes before output

So domain$\to$input$\text{; }$range$\to$ output

domain$\to x \in \left(- \infty , + \infty\right)$
range $\to y \in \left(- \infty , + \infty\right)$

Jun 14, 2017

Domain: $x \in \left(- \infty , \infty\right)$
Range: $y \in \left[- 8 , - \infty\right)$

Explanation:

The domain includes all the values of $x$ that you can "legally" plug into the equation. If plugging in zero forced you to divide by zero or put a negative into a square root, those would be outside the domain of acceptable $x$ values.

In the equation $y = - 2 {x}^{2} - 8$, you are neither dividing by an $x$ term, nor are you dealing with $x$ values under a radical sign. Therefore, the domain includes all real numbers. That means you can plug in any real number you like, from $- \infty$ to $+ \infty$.

Domain: All real numbers, $\mathbb{R}$, which can be written as:

$x \in \left(- \infty , \infty\right)$

The range is all the values of $y$ that this function can produce. Because the equation is a quadratic equation, we have a parabola. That means that the maximum of the parabola is the upper bound of the equation and there is no lower bound.

The vertex will be the upper most top part of this function.

Vertex: $x = - \frac{b}{2 a} = - \frac{0}{2 \left(- 2\right)} = \frac{0}{4} = 0$

Plugging $0$ into the function gives $y = - 8$. So the range is

Range: All values of $y$ less than $- 8$, or

$y \in \left[- 8 , - \infty\right)$