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### Course: Algebra (all content)>Unit 7

Lesson 24: Determining the range of a function (Algebra 2 level)

Learn how you can find the range of any quadratic function from its vertex form.
In other words, we will learn how to determine the set of all possible outputs of a given quadratic function.

## Let's study an example problem

We want to find the range of the function $f\left(x\right)=-2\left(x+3{\right)}^{2}+7$.
In this article, just as we're used to referring to inputs of a function with the letter $x$, we will refer to the outputs of a function with the letter $y$. For instance, $y=7$ is the output of $f$ for an input of $x=-3$ (this is just another way of saying $f\left(-3\right)=7$).
Finding the range of a function, just by looking at its formula, is pretty difficult! Actually, it's not even that easy to tell whether a single specific value is a possible output!
For instance, is $y=9$ a possible output of $f$?
In order to answer that question, we need to substitute $f$'s formula into $f\left(x\right)=9$ and solve. If we find a solution, then $y=9$ is a possible output. Otherwise, it isn't.
However, it's not possible to perform this check for every possible output, because they are infinite! This article will show two possible solution methods to work around this problem.

## Solution method 1: The graphical approach

It turns out graphs are really useful in studying the range of a function. Fortunately, we are pretty skilled at graphing quadratic functions.
Here is the graph of $y=f\left(x\right)$.
Now it's clearly visible that $y=9$ is not a possible output, since the graph never intersects the line $y=9$.
Let's perform similar checks for a couple more $y$-values.
Question 1Question 2
Is $y=-5$ a possible output of $f$?

Is $y=-50$ a possible output of $f$?

So we saw how we can check whether a given value is a possible output using a graph. A graph can actually tell us the entire range of possible outputs!
For instance, the graph of $y=f\left(x\right)$ shows that $7$ (the $y$-coordinate of the vertex) is the maximum $y$-value that the function outputs. Furthermore, since the parabola opens down, every $y$-value below $7$ is also a possible output.
In other words, the range of $f$ is all $y$-values less than or equal to $7$. This is it! Mathematically, we can write the range of $f$ as .

Consider the function $g\left(x\right)=\left(x-4{\right)}^{2}-5$ which is graphed below.
What is the range of $g$ ?

## Solution method 2: The algebraic approach

At this point, you may ask yourselves, "Do we always have to draw the graph when we want to find the range?" and you will be right in doing so! Laziness is a great motivation for finding better ways to solve problems.
Let's think about the work we did above and look for a pattern.
Our first function, $f\left(x\right)=-2\left(x+3{\right)}^{2}+7$, had a parabola that opened $\text{down}$ and whose vertex was at $y=7$. In consequence, its range was all $y$-values $\text{less}$ than or equal to $7$.
Our second function, $g\left(x\right)=\left(x-4{\right)}^{2}-5$, had a parabola that opened $\text{up}$ and whose vertex was at $y=-5$. In consequence, its range was all $y$-values $\text{greater}$ than or equal to $-5$.
It turns out all we need to know in order to determine the range of a quadratic function is the $y$-value of the vertex of its graph, and whether it opens up or down.
This is easy to tell from a quadratic function's vertex form, $y=a\left(x-h{\right)}^{2}+k$. In this form, the vertex is at $y=k$, and the parabola opens $\text{up}$ when $a>0$ and $\text{down}$ when $a<0$.

Use what you've learned to find the range of $h\left(x\right)=\frac{1}{2}\left(x-3{\right)}^{2}+2$.
$\left\{y\in \mathbb{R}\phantom{\rule{0.167em}{0ex}}|\phantom{\rule{0.167em}{0ex}}$
$\right\}$

## Want to join the conversation?

• How does the pattern work? Like how is the minimum or maximum value dependent on the constant term k in a(x+h)^2+k? Is it something to do with (x+h)^2 having its least value as 0?
• Yes. Since (x + h)^2 is a squared term, it is always positive and the smallest it can be is, as you say, zero. However, the "a" term plays a role too, since it determines whether the parabola is pointed up or down (whether you are looking for a minimum or a maximum). Then the "k" value gives you the number.
• I understand finding the range of a function, but I'm still confused about how to find the domain. Using the first function from the example, what would be the domain and how would you find it?
• Well, you see the first function goes on forever, right? So, it has an infinite amount of x-values, making the domain "All Real Numbers". All regular quadratic functions have this domain.
• How does one comes to realization that a function is not defined for a specific value of x so that then is not included in the domain? for example, I was given this: 1/(x-1)(x+2), but the answer came out that the function is not defined when x=1. and there were two domains for the function then.
• No, there are no "two" domains. It was the same domain of "all real numbers". But, look--in the function, (x-1)(x+2) was in the Denominator. We know that the denominator can't be zero, or else it would be undefined. So, we have to find values which could make the denominator zero, and specify it in the domain.
i.e., (x-1)(x+2) ≠ 0==> x ≠ 1 or x ≠ -2 (those values make the den. zero)
So, the domain is {x ∈ ℝ | x ≠ 1 or x ≠ -2} which just means that the domain contains all real numbers except 1 and -2, which makes the output undefined (1⁄0).

Hope that makes it clear and if I've made any mistake, please let me know. :)
• A quadratic function is given by y=5x^2+7x+11. For what domain of values of y, the solution of x will be a real number?
• I used the quadratic formula for that function and that results in an imaginary number for x.
• How am I supposed to enter the symbol for greater than, I thought it was supposed to be >- but the answer block would not accept it.
• it's not >- but >= with an equal sign, not a hyphen
• When do you know if it is less than AND equal to (≤) or only less than(<)? Basically it is a case of ≤ vs. <
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