SAT functions have the dubious honor of being one of the trickiest topics on the SAT math section. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics.
This means that the difference between missing points on this seemingly tricky topic and acing them is simply a matter of practice and familiarization. And considering that function problems generally show up on average of three to four times per test, you will be able to pick up several more SAT math points once you know the rules and workings of functions.
This will be your complete guide to SAT functions. We’ll walk you through exactly what functions mean, how to use, manipulate, and identify them, and exactly what kind of function problems you’ll see on the SAT.
What Are Functions and How Do They Work?
Functions are a way to describe the relationship between inputs and outputs, whether in graph form or equation form. It may help to think of functions like an assembly line or like a recipe--input eggs, butter, and flour, and the output is a cake.
Most often you’ll see functions written as $f(x) = \an \equation$, wherein the equation can be as complex as a multivariable expression or as simple as an integer.
Examples of functions:
$f(x) = 6$
$f(x) = 5x - 12$
$f(x) = x^2 + 2x - 4$
Functions can always be graphed and different kinds of functions will produce different looking graphs. On a standard coordinate graph with axes of $x$ and $y$, the input of the graph will be the $x$ value and the output will be the $y$ value.
Each input ($x$ value) can produce only one output, but one output can have multiple inputs. In other words, multiple inputs may produce the same output.
One way to remember this is that you can have “many to one” (many inputs to one output), but NOT “one to many” (one input to many outputs).
This means that a function graph can have potentially many x-intercepts, but only one y-intercept. (Why? Because when the input is $x = 0$, there can only be one output, or $y$ value.)
A function with multiple x-intercepts.
You can always test whether a graph is a function graph using this understanding of inputs to outputs. If you use the “vertical line test,” you can see when a graph is a function or not, as a function graph will NOT hit more than one point on any vertical line.
No matter where we draw a vertical line on our function, it will only intersect with the graph a maximum of one time.
The vertical line test applies to every type of function, no matter how "odd" looking.
Even "strange-looking" functions will always pass the vertical line test.
But any graph that fails the vertical line test (by intersecting with the vertical line more than once) is automatically NOT a function.
This graph is NOT a function, as it fails the vertical line test.
Too many obsticals in the way of the ascent works out as well for functions as it does for real life (which is to say: not well at all.)
Function Terms and Definitions
Now that we’ve seen what functions do, let’s talk about the pieces of a function.
Functions are presented either by their equations, their tables, or by their graphs (called the “graph of the function”). Let’s look at a sample function equation and break it down into its components.
An example of a function:
f(x) = x^2 + 5
$f$ is the name of the function
(Note: we can call our function other names than $f$. This function is called $f$, but you may see functions written as $h(x)$, $g(x)$, $r(x)$, or anything else.)
$(x)$ is the input
(Note: in this case our input is called $x$, but we can call our input anything. $f(q)$ or $f(\strawberries)$ are both functions with the inputs of $q$ and $\strawberries$, respectively.)
$x^2 + 5$ gives us the output once we plug in the input value of $x$
An ordered pair is the coupling of a particular input with its output for any given function. So for the example function $f(x) = x^2 + 5$, with an input of 3, we can have an ordered pair of:
$f(x) = x^2 + 5$
$f(3) = 3^2 + 5$
$f(3) = 9 + 5$
$f(3) = 14$
So our ordered pair is (3, 14).
Ordered pairs also act as coordinates, so we can use them to graph our function.
Now that we understand our function ingredients, let's see how we can put them together.
Different Types of Functions
We saw before that functions can have all sorts of different equations for their output. Let’s look at how these equations shape their corresponding graphs.
A linear function makes a graph of a straight line. This means that, if you have a variable on the output side of the function, it cannot be raised to a power higher than 1.
Why is this true? Because $x^2$ can give you a single output for two different inputs of $x$. Both $-3^2$ and $3^2$ equal 9, which means the graph cannot be a straight line.
Examples of linear functions:
$f(x) = x - 12$
$f(x) = 4$
$f(x) = 6x + 40$
A quadratic function makes a graph of a parabola, which means it is a graph that curves to open either up or down. It also means that our output variable will always be squared.
The reason our variable must be squared (not cubed, not taken to the power of 1, etc.) is for the same reason that a linear function cannot be squared--because two input values can be squared to produce the same output.
For example, remember that $3^2$ and $-3^2$ both equal 9. Thus we have two input values--a positive and a negative--that give us the same output value. This gives us our curve.
(Note: a parabola CANNOT open side to side because it would have to cross the y-axis more than once. This, as we’ve already established, would mean it was not a function.)
This is NOT a quadratic function, as it fails the vertical line test.
A quadratic function is often written as:
$f(x) = ax^2 + bx + c$
The $a$ value tells us how the parabola is shaped and the direction in which it opens.
A positive $a$ gives us a parabola that opens upwards
A negative $a$ gives us a parabola that opens downwards
A large $a$ value gives us a skinny parabola
A small $a$ value gives us a wide parabola
The $b$ value tells us where the vertex of the parabola is, left or right of the origin.
A positive $b$ puts the vertex of the parabola left of the origin
A negative $b$ puts the vertex of the parabola right of the origin
The $c$ value gives us the y-intercept of the parabola. This is wherever the graph hits the y-axis (and will only ever be one point).
(Note: when $b = 0$, the y-intercept will also be the location of the vertex of the parabola.)
Don't worry if this seems like a lot to memorize right now--with practice, understanding function problems and their components will become second nature.
Typical Function Problems
SAT function problems will always test you on whether or not you properly understand the relationship between inputs and outputs. These questions will generally fall into four question types:
- Functions with given equations
- Functions with graphs
- Functions with tables
- Nested functions
There may be some overlap between the three categories, but these are the main themes you’ll be tested on when it comes to functions. Let’s look at some real SAT math examples of each type.
A function equation problem will give you a function in equation form and then ask you to use one or more inputs to find the output (or elements of the output).
In order to find a particular output, we must plug in our given input for $x$ into our equation (the output).
So if we want to find $f(2)$ for the equation $f(x) = x + 3$, we would plug in 2 for $x$.
$f(x) = x + 3$
$f(2) = 2 + 3$
$f(2) = 5$
So, when our input ($x$) is 2, our output ($y$) is 5.
Now let’s look at a real SAT example of this type:
As you can see here, we are given our equation as well as two inputs and their combined output. We must use this knowledge to find an element of our output (in this case, the value of $a$.)
So let us find our outputs for each input we are given.
$f(x) = 5x - 2a$
$f(10) = 5(10) - 2a$
$f(10) = 50 - 2a$
$f(x) = 5x - 2a$
$f(5) = 5(5) - 2a$
$f(5) = 25 - 2a$
Now, let us set the sum of our two outputs equal to 55 (as was stipulated in the question).
$50 - 2a + 25 - 2a = 55$
$75 - 4a = 55$
$-4a = -20$
$a = 5$
Our final answer is C, $a = 5$.
A function graph question will provide you with an already graphed function and ask you any number of questions about it.
These questions will generally ask you to identify specific elements of the graph or have you find the equation of the function from the graph.
So long as you understand that $x$ is your input and that your equation is your output, $y$, then these types of questions will not be as tricky as they appear.
We know that $f(x)$ is the output or $y$ value. So let us reimagine that this question is asking you: “What are the values of x for which y is positive?”
We can see that the y value is positive (more than 0) between the points (1, 0) and (4, 0) and between the points (8, 0) and (10, t).
This means that $y$ (or $f(x)$) will be positive for any values of $x$ between 1 and 4 and between 8 and 10.
So our final answer is C, 1 < x < 4 or 8 < x ≦ 10.
The third way you may see a function is in its table.
You will be given a table of values both for the input and the output and then asked to either find the equation of the function or the graph of the function.
Instead of using $x$ for our input, this problem has us use $t$. If you become very used to using $f(x)$, this may seem disorienting, so you can always rewrite the problem using $x$ in place of $t$.
In this case, we will continue to use $t$, just so that we can keep the problem organized on the page.
First, let us find the y-intercept.
The y-intercept is the point at which $x = 0$, so we can see that we are already given this with the first set of numbers in the table. When $t = 0$, $f(t) = -1$
Our y-intercept is therefore -1, which means that we can automatically eliminate answer choices B, C, and E.
Now, let us use the strategy of plugging in answers to make our lives simpler. This way, we don’t have to actually find the equation on our own--we can simply test which answer choices match the inputs and outputs we are given in our table.
Our answer choices are between A and D, so let us first test A with the second ordered pair.
Our potential equation is:
$f(t) = t - 1$
And our ordered pair is:
So let us put them together.
$f(t) = t - 1$
$f(1) = 1 - 1$
$f(1) = 0$
This is incorrect, as it would mean that our output is 0 when our input is 1, and yet the ordered pair says that our output will be 1 when our input is 1.
Answer choice A is incorrect.
By process of elimination, let us try answer choice D.
Our potential equation is:
$f(t) = 2t - 1$
And our ordered pair is again:
So let us put them together.
$f(1) = 2(1) - 1$
$f(1) = 2 - 1$
$f(1) = 1$
This matches the input and output we are given in our ordered pair. Answer choice D is correct.
(Note: it is generally a good idea to test more than one ordered pair, as two equations may occasionally get the same ordered pair. In this case, we stopped here as there were no other answer choices that could possibly match.)
Our final answer is D, $f(t) = 2t - 1$
The final type of function problem you might encounter on the SAT is called a “nested” function. Basically, this is an equation within an equation.
In order to solve these types of questions, think of them in terms of your order of operations. You must always work from the inside out, so you must first find the output for your innermost function.
Once you’ve found the output of your innermost function, you can use that result as the input of the outer function.
Let’s look at this in action to make more sense of this process.
What is $f(g(x - 2))$ when $f(x) = x^2 - 6$ and $g(x) = 3x + 4$?
Because $g(x)$ is nested the deepest, we must find its output before we can find $f(g(x - 2))$.
Instead of a number for $x$, we are given another equation. Though this may look different from earlier problems, the principle is exactly the same--replace whatever input we have for the variable in the output equation.
$g(x) = 3x + 4$
$g(x - 2) = 3(x - 2) + 4$
$g(x - 2) = 3x - 6 + 4$
$g(x - 2) = 3x - 2$
So our output of $g(x - 2)$ is $3x - 2$. Again, this is an equation and not an integer, but it still works as an output.
Now we must finish the problem by using this output of $g(x)$ as the input of $f(x)$.
(Why do we do this? Because we are finding $f(g(x))$, which positions the result/output of $g(x)$ as the input of $f(x)$.)
$f(x) = x^2 - 6$
$f(g(x - 2)) = (3x - 2)^2 - 6$
Now, we have a bit of a complication here in that we must square an equation. If you remember your exponent rules, you know you cannot simply distribute the square across the elements of the equation; you must square the entire expression.
So let’s take a moment to expand $(3x - 2)^2$ before we find the solution for the entire equation.
$(3x - 2)^2$
$(3x - 2)(3x - 2)$
$(3x * 3x) + (3x * -2) + (-2 * 3x) + (-2 * -2)$
$9x^2 -6x - 6x + 4$
$9x^2 - 12x + 4$
Now, let us add this expanded form of the equation back into the output.
$f(g(x - 2)) = (9x^2 - 12x + 4) - 6$
$f(g(x - 2)) = 9x^2 - 12x - 2$
So our final solution for $f(g(x - 2))$ is $9x^2 - 12x - 2$.
Our final answer is E, $9x^2 - 12x - 2$.
Functions within functions, dreams within dreams. Make sure not to lose yourself along the way.
Strategies for Solving Function Problems
Now that you’ve seen all the different kinds of function problems in action, let’s look at some tips and strategies for solving function problems of various types.
For clarity, we’ve split these strategies into multiple sections--tips for all function problems and tips for function problems by type. So let’s look at each strategy.
Strategies for All Function Problems:
1) Keep careful track of all your pieces and write everything down
Though it may seem obvious, in the heat of the moment it can be far too easy to confuse your negatives and positives or misplace which piece of your function (or graph or table) is your input and which is your output. Parenthesis are crucial.
The creators of the SAT know how easy it is to get pieces of your function equations confused and mixed around (especially when your input is also an equation), so keep a sharp eye on all your moving pieces and don’t try to do function problems in your head.
2) Use PIA and PIN as necessary
As we saw in our function table problem above, it can save a good deal of effort and energy to use the strategy of plugging in answers. You can also use the technique of plugging in your own numbers to test out points on function graphs, work with any variable function equation, or work with nested functions with variables.
For instance, let’s look at our earlier nested function problem using PIN. (Remember--most any time a problem has variables in the answer choices, you can use PIN).
What is $f(g(x - 2))$ when $f(x) = x^2 - 6$ and $g(x) = 3x + 4$?
So let us say that the $x$ is the $g(x - 2)$ function is 5. (Why 5? Why not!)
Now $x - 2$ will be $5 - 3$, or 3. This means $g(x - 2)$ will be $g(3)$.
$g(x - 2) = 3x + 4$
$g(3) = 3(3) + 4$
$g(3) = 9 + 4$
$g(3) = 13$
Now, let us plug this number as the value for our $g(x - 2)$ function into our nested function $f(g(x - 2))$.
$f(x) = x^2 - 6$
$f(g(3)) = (13)^2 - 6$
$f(g(3)) = 169 - 6$
f(g(3)) = 163$
Finally, let us test our answer choices to see which one matches our found answer of 163.
Let us, as usual when using PIA or PIN, start in the middle with answer choice C.
$9x^2 - 24x + 10$
Now, we replace our $x$ value with the $x$ value we chose originally--5.
$9x^2 - 24x + 10$
$9(5)^2 - 24(5) + 10$
$9(25) - 120 +10$
$225 - 120 + 10$
Unfortunately, this number is too small. Let us try answer choice D instead.
$9x^2 - 12x + 4$
$9(5)^2 - 12(5) + 4$
$9(25) - 60 + 4$
$225 - 60 + 4$
$165 + 4$
This value is still too large, but we can see that it is awfully close to the final answer we want. Just by looking over our answer choices, we can see that answer choice E is exactly the same expression as answer choice D, except for the final integer value. If we were to subtract 2 from 165 instead of adding 4 (as we did with answer choice D), we would get our final answer of 163.
As you can see.
$9x^2 - 12x - 2$
$9(5)^2 - 12(5) - 2$
$9(25) - 60 - 2$
$225 - 60 - 2$
$165 - 2$
So our final answer is E, $9x^2 - 12x - 2$.
3) Practice, practice, practice
Finally, the only way to get truly comfortable with any math topic is to practice as many different kinds of questions on that topic as you can. If functions are a weak area for you, then be sure to seek out more practice questions.
For Function Graphs and Tables:
1) Start by finding the y-intercept
Generally, the easiest place to begin when working with function graphs and tables is by finding the y-intercept. From there, you can often eliminate several different answer choices that do not match our graph or our equation (as we did in our earlier examples).
The y-intercept is almost always the easiest piece to find, so it’s always a good place to begin.
2) Test your equation against multiple ordered pairs
It is always a good idea to find two or more points (ordered pairs) of your functions and test them against a potential function equation. Sometimes one ordered pair works for your graph and a second does not.
You must match the equation to the graph (or the equation to the table) that works for every coordinate point/ordered pair, not just one or two.
For Function Equations and Nested Equations:
1) Always work inside out
Nested functions can look beastly and difficult, but take them piece by piece. Work out the equation in the center and then build outwards slowly, so as not to get any of your variables or equations mixed up.
2) Remember to FOIL
It is quite common for SAT to make you square an equation. This is because many students get these types of questions wrong and distribute their exponents instead of squaring the entire expression.If you don’t properly FOIL, then you will get these questions wrong. Whenever possible, try not to let yourself lose points due to these kinds of careless errors.
For instance, let's say that you must square an expression.
Sqaure the expression $x + 3$.
We are told to square the entire expression, so we would say:
$(x + 3)^2$
Now you must FOIL this out properly.
$(x + 3)(x + 3)$
$(x * x) + (3 * x) + (3 * x) + (3 * 3)$
$x^2 + 3x + 3x + 9$
$x^2 + 6x + 9$
The final expression, once you have squared $x + 3$, is:
$x^2 + 6x + 9$.
(Note: It is a common error for students to distribute the square and say: $(x + 3)^2 = x^2 + 9$ but this is WRONG. Do not fall into this kind of trap!)
You're all leveled-up--time to fight the big boss and put knowledge to action!
Test Your Knowledge
Now let's put your function knowledge to the test against real SAT math problems.
Answers: A, E, B, C
1) As per our strategies, we will start by finding the y-intercept. We can see in this graph that the y-intercept is +2, which means we can eliminate answer choices C and E. (Why did we eliminate answer choice E? Because it had no y-intercept, which means that its y-intercept would be 0).
We can see that the vertex of the graph is at $x = 0$ and so it is not shifted to the right or left of the y-axis. This means that, in our quadratic equation $ax^2 + bx + c$, our $b$ value has to be 0. If it were anything other than 0, our graph would be shifted left or right of the y-axis.
Now answer choices B and D are squaring expressions, so let us properly FOIL them in order to see the equation properly.
Answer choice B gives us:
$y = (x + 2)^2$
$y = (x + 2)(x + 2)$
$y = x^2 + 2x + 2x + 4$
$y = x^2 + 4x + 4$
This equation would give us a parabola whose y-intercept was at +4 and whose vertex was positioned to the left of the y-axis (remember, a positive $b$ value shifts the graph to the left.)
We can eliminate answer choice B.
By the same token, we can also eliminate answer choice D, as it would give us:
$y = (x - 2)^2$
$y = (x - 2)(x - 2)$
$y = x^2 - 4x + 4$
Which would give us a graph with a y-intercept at +4 and a vertex positioned to the right of the y-axis.
By process of elimination, we are left with answer choice A. But, for the sake of double-checking, let us test a coordinate point on the graph against the formula.
We already know that our equation matches the coordinate points of (0, 2), as that is our y-intercept, but there are several more places on the graph that hit at even coordinates.
By looking at the graph, we can see that the parabola hits the coordinates (1, 3), so let us test this point by plugging our input (1) into our equation, in hopes that it will match our output of 3.
$y = x^2 + 2$
$y = (1)^2 + 2$
$y = 1 + 3$
$y = 3$
Our equation matches two sets of ordered pairs on the graph. We can reasonably say that this is the correct equation for the graph.
Our final solution is A, $y = x^2 + 2$
2) This problem might seem tricky because we are not actually changing our input, but instead are altering our output. This is because the input (in the parentheses) remains unchanged, while we are doubling the entire output and adding 4.
So if we start with:
$f(x) = 3x + 4$
And we must transform it into:
$2f(x) + 4$
Then we would multiply the output by 2 and add 4, like so:
$2(3x + 4) + 4$
$6x + 8 + 4$
$6x + 12$
Though the math isn’t difficult in this problem, it is tricky in that you are more used to transforming the input, not the output. Always keep close track of how the question is asking you to shift your parameters.
Our final answer is E, $6x + 12$
3) Again, let us begin by finding our y-intercept. We know the y-intercept is the value of y at which point $x = 0$, which in this case is 3.
This means we can eliminate answer choices C, D, and E straight away, as only A and B have equations with y-intercepts at +3.
Now let us test one of our coordinate points against both equations A and B.
One of our ordered pairs/coordinates is at (2, 9), so let us plug in 2 for our input in each equation and see which one gives us an output of 9.
Answer choice A gives us the equation:
$x^3 + 3$
So when we plug in our input, we get:
$2^3 + 3$
$8 + 3$
This does not match our output of 9, so we can eliminate answer choice A. (Note: there is no need to test any other answer options--the equation must work for all the ordered pairs, so one ordered pair wrong means that the entire equation must be wrong.)
By process of elimination, we are left with answer choice B, but let us double-check the same ordered pair, (2, 9), just to be sure:
$3x + 3$
$3(2) + 3$
$6 + 3$
Success! We have found our correct equation and eliminated all other possible answers.
Our final answer is B, $y = 3x + 3$
4) This question requires us to plug in a new input. The only tricky part of this is that our new input is an equation, so we must keep it intact until we can distribute it out.
$f(x) = 5x$
$f(p + r) = 5(p + r)$
Now, we can distribute the equation properly.
$f(p + r) = 5p + 5r$
Our final answer is C, $5p + 5r$
You did it! High fives all around.
The Take Aways
Many students have not dealt a lot with functions, but don’t let these kinds of questions intimidate or confuse you when you see them on the SAT. The principles behind functions are a simple matter of input, output, and plugging in values. The test will try to muddy the waters when they can, but always remember that these questions will appear to be more complex than they truly are.
Though it can be easy to make a error with your signs or variables, the actual problems are simple at their core. So pay close attention, double-check your work, and you’ll soon be able to work through functions problems with little trouble.
Phew! Knowing your functions means knowing a significant portion of the SAT math section (round of applause to you!), but there are so many more topics to cover. Take a look at all the topics you'll be tested on in the SAT math section and then mosey on over to our math guides to review any topic you feel rusty on. Not feeling confident about your exponent rules? How about your understanding of polygons? Need to review your slopes? Whatever the topic, we've got you covered!
Running out of time on the SAT math section? Not to worry! We have the tools and strategies to help you beat the clock and maximize your point gain.
Trying for a perfect score? Check out how to push your score to its maximum potential with our guide to getting an 800 on the SAT math, written by a perfect-scorer.
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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.