Not all function questions have weird symbols, some are just vanilla f(x) type things. You’ve probably been working with the f(x) notation in school for some time now, but let’s review some of the things you’ll see over and over again on the SAT:

##### Interpreting function notation

One thing you’re definitely going to need to be able to do is interpret function notation. For some questions, it’s enough to remember that saying f(x) = x3, for example, is basically the same as saying y = x3.

For other questions, you’re going to need to take that a bit further and identify points on a graph using function notation. Here’s a quick cheat for you (with colors!). When you have f(x) = y, that’s the same as an ordered pair (xy). For example, if you know that f(4) = 5, then you know that the graph of the function f contains the point (4, 5). Likewise, if you know that h(c) = p, you know that the graph of function h contains the point (cp). And so on. Basically, whatever is inside the parentheses (that’s called the argument, if you care) is your x-value, and whatever’s across the equal sign is your y-value. This is important. If you don’t understand yet, read it over and over until you do. Might help to write it down. Just sayin’.

To make sure you’ve got this, think about what the following things mean. Once you’ve thought about them, hold your mouse over them (don’t click, just hover…follow directions) to see what they’re about.

• f(0) = 3
• p(12) = 0
• s(3) = r(3)
##### Nested functions (functions in functions)

Have you ever seen those dolls where you open them up and there are smaller ones inside? If you haven’t, I’ve put a picture of some really artsy-fartsy ones there on the right. They’re called Russian nesting dolls. Welcome to the world. Can I take your coat?

Anyway, sometimes the SAT will put a function inside another function to try to bamboozle you. Don’t let yourself get stymied here. All you need to do is follow the instructions, same as you do with all other function questions. Let’s look at an example:

1. If f(x) = x2 – 10 and g(x) = 2f(x) + 3, what is g(√2)?

(A) 7
(B) 5
(C) –5
(D) –7
(E) –13

Let’s just take this one step at a time. First, let’s take the √2 we’re given and put it in for every x we see in g(x).

g(√2) = 2f(√2) + 3

Not so bad, right? Now let’s replace the f(√2) with an expression we can actually work with. Remember that f(x) = x2 – 10, so we write:

g(√2) = 2[(√2)2 – 10] + 3

See how this is working? Now just simplify:

g(√2) = 2[–8] + 3
g(√2) = –16 + 3
g(√2) = –13

So our answer is (E). Awesome, right?

##### Interpreting Graphs and Tables Say the picture above shows the function g(x) over a given range. Note that, although we have no equation for g(x), we still know a lot about it. We know, for example, that g(4) = –2. We also know that the y-intercept, g(0), is about –1. Think about the following, again, mousing over them once you know what’s up to see if you’re right:

• If g(a) = 0, how many possible values are there for a in the given range?
• When, in the given range, is g(x) < –4?

You could also be presented with function information in table form. Peep this:

 x 2 3 4 5 6 f(x) 13 18 25 34 45

Just like in the graph above, we can use this table to find points. For example, f(5) = 34. See if you can do the following:

• If f(p) = 18, p =
• What is f(6 – 4)?
• What is f(6) – f(4)?
• Holy crap those are different?
• Can you figure out what the function f(x) is?
##### Graph Translation

Sometimes the SAT likes to test you on whether you can figure out where a graph will move based on some manipulation of its equation. Usually, though, they won’t give you the equation. They’ll draw some crappy squiggly like the one above, call it g(x), and then ask what will happen to g(x+1).

I’ll give you the rules for this, but I highly recommend reminding yourself of them with your calculator if you should need them on your SAT. It’s very easy to set a simple function (like f(x) = x2, which you’ve seen a million times) as your starting point, and experiment with your graphing calculator to see how graphs will behave based on various modifications. In fact, why don’t you play with this widget a bit right now to see what happens?

Here are those rules:

• f(x)+1 ⇒ (graph moves UP one)
• f(x)–1 ⇒ (graph moves DOWN one)
• f(x+1) ⇒ (graph moves LEFT one)
• f(x–1) ⇒ (graph moves RIGHT one)
##### Sample Questions!

You need to be registered and logged in to take this quiz. Log in or Register kelly nguyen says:

This is really helpful!. This was just as I needed for the SAT that I am taking next year. Through your explanation on how to solve these types of problems, I was not scared of functions at all and they are really easy to understand! Agin thanks really much. Mike McClenathan says:

You’re welcome Kelly; it’s my pleasure. Shaane says:

How do u do #16?
If f(p) = 18, p =What is f(6-4)?What is f(6) – f(4)? Mike McClenathan says:

When a question says something like f(b) = g(b), that means the graphs of f and g intersect when xb.

This is actually really important, so don’t just breeze by it if it’s not obvious. Make sure you really get this.
Since the graph shows the two functions intersecting when x is negative, the only choice that could possibly be b is -3. Mike McClenathan says:

It’s really helpful to think of function notation as a fancy way of writing an ordered pair. For example, f(3) = 2 is just a fancy way of saying that on the graph of the function f, you’ll find the point (3, 2).

Using the table, I see that f(x) = 18 when x = 3. So since f(p) = 18, p = 3.

f(6-4) is going to be the same as f(2), since you always resolve what’s inside the parentheses first. That’s 13.

f(6) – f(4) = 45 – 25 = 20. Stacey Howe-Lott says:

Why does the graph move in the opposite direction on the x axis? Mike McClenathan says:

Good question — this is one of the most confusing things about graph translation. If you’re a numbers person, then look at the numbers (and plot them as you go):

When x = 0:
x^2 = 0
(x+1)^2 = 1

When x = 1:
x^2 = 1
(x+1)^2 = 4

…and so on.

If you don’t love that, then try this:

Don’t think of it like the graph is moving backwards. Think of it like a race along the x-axis from left to right. f(x + 1) gets where it’s going FASTER than f(x). It always stays 1 step ahead. Stacey Howe-Lott says:

Yeah – I don’t love the math explanation 🙂  I think I’m going to like the race one -but it still seems backwards to me.  Wouldn’t the race be from right to left – b/c f(x+1) is going to be switched one unit to the left from the graph of f(x)? Mike McClenathan says:

OK, try this: think of the x-axis like a timeline, where f(x + 1) is one step ahead, showing f(x) what to do. Like it’s teaching it a dance. First f(x + 1) does something, and then f(x) does it one second later. Mike McClenathan says:

One more (less mathy) thought. You could call it a race and f(x) is winning because f(x + 1) is heavier? That way it fits the left-to-right paradigm nicely? Stacey Howe-Lott says:

🙂  I’m loving the analogies.  But the thing I can’t get around is +1 always means move one to the RIGHT on a timeline, numberline, or an x-axis.

The thing plus one is always heavier, faster, or leading the pack Mike McClenathan says:

Yes, BUT, it’s not the function that’s being added to here. It’s the argument that goes INTO the function. Another analogy that’s a real stretch but I’m hungry:

Imagine trying to toast a frozen bagel. The bagel is x, and toaster is f(x). It takes a long time to toast a bagel when you pull it right out of the freezer. Now pretend the bagel is thawed already. It should take it less time to toast, but you’re going to put it through the same function: the toaster function. The thawed bagel is (x + 1). If you imagine the graph of toastedness, over time, which one will rise faster (which curve will be further to the left)? The f(x + 1) curve. Because what’s being fed into the function is already ahead. Stacey Howe-Lott says:

Oh – the FUNCTION (head-smack!)  Now I totally see the dancing – f(x+1) is always going to do the move first – in order to do it first – they have to scoot one to the left so that will be the first curve we see.  BRILLIANT! Mike McClenathan says:

This was really fun, and a great exercise in rejiggering an explanation until it clicks (something all tutors need to practice constantly lest their methods ossify).

Thank you for asking such good questions. I’ll be referring back to this conversation in my tutoring for years to come. 🙂 Stacey Howe-Lott says:

😉  Gotta lead by example – “there are no dumb questions!”

So yes, keeping your mind open and flexible so you can see problems through their eyes and understand where and what they are missing – very useful! Stacey Howe-Lott says:

Why does it move the opposite way on the x axis? Sally says:

Can you please explain #20? i don’t know how the answer is 15? thank you! Mike McClenathan says:

To do number 20, first figure out what g(7) is. According to the table, g(7) = -3. Since the f function is just the absolute value of the g function, that makes f(7) = 3, which means t = 3.. From there, you look at g(3), which is 15. Since it’s already positive, the absolute value brackets don’t change it, and f(3) is also 15. A314 says:

Thank you so much, this is very helpful 🙂 Mike McClenathan says:

You’re welcome! ddwad says:

I really need help with this question!
If f(x)=x2 and g(x)= x+3, then g(f(x))=

and this one:
If f(x)= x+1, then 1/f(x) X f(1/x)= Mike McClenathan says:

The first question is just a substitution question dressed up to look much harder than it is. What if I worded it this way:

p = x^2 and g(x) = x + 3. What is g(p)?

It’d be a little easier, right?

Back to the question as it’s asked. Work backwards:

g(f(x)) = g(x^2) ← because f(x) = x^2
g(x^2) = x^2 + 3 ← because g(x) = x + 3 Mike McClenathan says:

The second question is similar. Take the pieces one at a time.

We can do simple substitution for the first part. Because f(x) = x + 1, 1/f(x) is just 1/(x + 1).

For the other part, f(1/x), we just put (1/x) into the function everywhere x was there before. Not so bad in this case, since the function is fairly simple. If f(x) = x + 1, then f(1/x) = (1/x) + 1.

Now all we need to do is put them together and simplify!

[1/(x + 1)]×[(1/x) + 1]

At this point, you can simplify as much or as little as you’d like, depending on what your answer choices are.

It’s likely the answer choice that you want is 1/x. It’s tricky algebra, but you can simplify to that. Wanna show me how? DDwad says:

Don’t I have to find the common denominator for (1/x) + 1? This is the part where I’m having the most trouble with. 🙁 Mike McClenathan says:

What if you take the fraction (1/x + 1)/(x + 1) and multiply it by x/x? 🙂 Yuper says:

If f(x)= (x-1)^2+1, what is the y intercept of te graph of f(x+1) in the xy plane? Mike McClenathan says:

Well, f(x + 1) is going to be ((x + 1) – 1)^2 + 1, or just x^2 + 1.That has a y-intercept of 1.

Note that if you foil out first, this question becomes much more of a pain in the butt. 🙂 Bleach says:

I need help with function questions from the blue book. 🙁
Pg. 312 #16 pg. 313 #18 and pg.338 #10. Can you please try to explain it as simply stupid as possible because I don’t understand any other methods. Thanks! Mike McClenathan says:

I can explain these, but for now let me just give you some hints. I wouldn’t actually categorize them as function difficulties—the first one is an absolute values question, and the second one is an exponents question. #10 on pg 338 is also an exponents question, which I coincidentally just answered on the Q&A page earlier today. If you’re still stuck on them after wrestling a little more with them, please submit them to me at the Q&A page.

The first one, #16, is a slightly novel version of a classic absolute value question. Read this post, and see if it helps you. Note that in #16, they ask you for the inequality that gives all the pumpkins he doesn’t want, which is why the answers have greater than signs instead of less than signs.

For #18, remember what a negative exponent means. The second part of the expression (x^(-2)) is just going to get tinier and tinier as x gets bigger and bigger. Bleach says:

Oh… I see for #16. I just need to plug in some #s that Mr.Sephera isn’t using and see which answer choice will be true no matter what, right?

And for #18. Since you said negative exponents get smaller and smaller each time, the answer is x^3 because once you subtract x^3 and (x)^-2, it will be closest to x^3 since x^-2 is so small, right? Mike McClenathan says:

Bingo. Nice work! Meaghan O says:

Could you please explain #7 on the Diagnostic 1? Thanks (: Mike McClenathan says:

Sure. Simply plug the values for x into the functions given in the answer choices, and eliminate the choices that don’t give you what the table says you’re supposed to get. Start with the first entry, which says that when x = 2, f(x) = 11.

(A) 8(2) – 7 = 9. That’s not 11, so eliminate choice (A).
(B) 5(2) + 1 = 11. So keep choice (B).
(C) -3(2) – 9 = -15. That’s not 11, so eliminate choice (C).
(D) (2)^2 + 7 = 11. So keep choice (D).
(E) 6(2) – 1 = 11. So keep choice (E).

Now try x = 3. You know f(3) is supposed to be 16. Note that you don’t need to try the choices you’ve already eliminated.

(B) 5(3) + 1 = 16. Yep. Still keep (B).
(D) (3)^2 + 7 = 16. Yep. Still keep (D).
(E) 6(3) – 1 = 17. Nope. Eliminate (E).

You shouldn’t be surprised that the question’s making you go all the way to the end here. That’s what makes it a good question. The last thing you know is that f(4) = 23.

(B) 5(4) + 1 = 21. That’s not 23. Eliminate (B).
(D) (4)^2 + 7 = 23. Yes indeed. (D) is your final answer. Brenna says:

I’m from Canada and I am going to write my SAT next month. I studied all summer for the English portion and decided to leave the math until last minute because I am a strong math student (99% in grade 10 academic math). I have been reading your blog and been trying to understand how to do the SAT math but it doesn’t click with my brain and the way I have been taught in Canada. When the letter f or another variable is put in front of an equation what is that letter supposed to represent? And in #20 what does that vertical bar represent? Mike McClenathan says:

When the letter f (or another letter) is placed in front of an equation like you see in this post, it’s not a variable at all–it’s denoting a function. So f(x) means “a function called ‘f’ of the variable ‘x’.” It does not mean “f times x.” So if f(x) = x + 3, then all that means is that f is a function that adds 3 to whatever is input to it. f(a) = a + 3, f(2) = 2 + 3, and f(10) = 10 + 3.

The vertical bar in #20 is for absolute value. Absolute value is tested on the SAT, but not very often. Functions, however, are tested all the time. You have plenty of time to learn them, but you should get started. Zelo says:

HELP! :0

WHat is the domain of the function f(x)=5/2x^2-x-3? Mike McClenathan says:

The SAT doesn’t really test domain and range, so I don’t want to get into it here. You might want to try Wolfram Alpha. Bill Shillito says:

I have in fact seen domain tested on the SAT before, if only in the simplistic sense of “don’t break the function.” How do you break a function? Two ways:

* Make the denominator equal zero. (This creates a black hole that will destroy the universe.)
* Make the inside of a square root negative. (This turns into imaginary numbers and soon you will need to see a psychiatrist.)

So, assuming that you meant 5 / (2x² – x – 3), you’d have to factor the polynomial in the denominator, and then set it equal to zero and solve for x. So your domain is everything but those x values you just found — they are what breaks the function. AA says:

Q17 drill #2? I have no idea why, but i’m stumped! AA says: Eliza says:

Could you explain problem 13? Mike McClenathan says:

Sure. If f(x) = 2x – 1, then

f(10) = 2(10) – 1 = 19
f(5) = 2(5) – 1 = 9

So f(10) – f(5) = 19 – 9 = 10 Lyubomir Gizdarski says:

On page 112 (#17) of your guide, when it says “Which of the following could be an expression of h(x) in terms of g(x),” how do you know if the question is asking you to describe the change from either h(x) to g(x) or from g(x) to h(x)? I’m getting confused with the meaning of the “in terms of ___” part. Mike McClenathan says:

Pro-tip: When you see “x in terms of y” that just means that the answer choices will contain possible expressions of x, all of which will contain ys (as opposed to numerical values for x). An illustration: say 3x + 9y = 30. To get x in terms of y, all we do is solve for x:

3x + 9y = 30
3x = 30 – 9y
x = 10 – 3y

Another way to think about this: the important part isn’t the “in terms of,” it’s what comes before it. That’s what tells you what to solve for.

So in the question you mention, the h function is clearly 3 units below the g function. You’re looking for an expression that says so. Choice (E) is the answer: h(x) = g(x) – 3. Liara Tsoni says:

I don’t know how you figured out what the function f(x) is 🙁 Mike McClenathan says:

🙂 Don’t worry about that—it’s just trial and error. You see that the rate of increase isn’t constant (the function gets bigger from 3 to 4 than it does from 2 to 3) so you know it’s not linear. From there, you just have to try some things and see if you get close, then try some other things and get closer.

You’ll never have to do this on the SAT. It was just for fun. Nick Twyman says:

You mislabeled Question 16 as 17 on the answer key. Mike McClenathan says:

Wow…sure enough! Thanks for your keen eyes—I’ll change it right away. trippy says:

wow, this is amazing. i am hooked on your website and am totally buying your book! amazing tips and really clear answers that are in ‘simple’ english:) vivienne uzor says:

pls could you explain the second one? Mike McClenathan says:

f(b) = g(b) means that the functions intersect when x = b. Since the graph shows us that the only intersection is when x is negative, the answer must be (A). Guest says:

Hi, could you please explain number one on diagnostic #2? thanks Mike McClenathan says:

That one’s explained in this comment on the Drill 2 post. The Ridgewood Tutor says:

“Can I take your coat”? This is why this is my fav prep site…