Here's a method without calculus.

Let Denominator can never be in this case. So the domain depends only upon $4y^2-24+4y\Rightarrow 4y^2+4y-24≥0\Rightarrow 4(y-2)(y+3)≥0\Rightarrow (y-2)(y+3)≥0\Rightarrow y\in(- \infty, -3]\cup[2,\infty).$

Hence the range of the is

I guess you know solving inequalities...

The **range **of the given **function **from the given graph are: all values of y that are greater than or equal to -3 . What is the Range of a Function? **Range **of a **function **are plotted on the vertical axis (y-axis), and is the set of y-values in the function . In the graph given, the y-values of the **function **on the y-axis are -3 and all values greater than -3. Therefore, the **range **of the given **function **from the given graph are: all values of y that are greater than or equal to -3 . Learn more about range of a function on: https://brainly.com/question/7644636

For anything other that straight lines (ie quadratics), you cannot assume that a function hits the maximum or minimum at the limits of the domain. Draw any arbitrary curve (going up and down). Now draw two vertical lines to show the domain. Can you see that there is no reason to believe that the range is determined by the domain end points? It is possible to do this without a graph? Yes, but that might require something like calculus for a complicated function. Most students would simply sketch one just to remind themselves what kind of function they are dealing with.

The question asks you to use a graph.

Have you tried looking at the graph of $y=f(x)$?

By examining the graph for our domain $-3 \leq x \leq 3$, it should be pretty clear that the range is $y \geq f(3)$ as the $y$ value never falls below $f(3)$ in our domain.

Remember that our function is $f(x) = \frac{1}{x^2}$, not $f(x) = {x^2}$. So $f(3)$ evaluates to $\frac{1}{9}$, not $9$.

Your range should also be given using $y$, not $x$, as the range represents a set of $y$ values.

Here's another way to think about it: consider what happens to $f(x)$ as $x$ approaches $0$. As $x$ becomes smaller, so does $x^2$. And fractions become larger in value as their denominators become smaller (for example, take $\frac{1}{0.01} = 100$). So as $x$ approaches $0$, $f(x)$ approaches infinity, i.e. $\lim_{x\to0} f(x) = \infty$.

The function $f(x) = 1/x$ has domain $D_f = (-\infty, 0) \cup (0, \infty)$ and range $R_f = (-\infty, 0) \cup (0, \infty)$.

Based on your work, we can see that the graph of the function $$g(x) = \frac{x - 1}{x + 2} = 1 - \frac{3}{x + 2}$$ is obtained from the graph of $f(x)$ by reflecting the graph of $f(x)$ in the $x$-axis, stretching it vertically by a factor of $3$, and translating it two units to the left and one unit up. Consequently, the domain of $g(x)$ is $D_g = (-\infty, -2) \cup (-2, \infty)$ and its range is $R_g = (-\infty, 1) \cup (1, \infty)$.