Advanced Math

Function Notation

We often wish to formulate a relationship between two quantities, such as how far a vehicle travels over how much time, or how much profit a restaurant will have based on its online ratings. Using these relationships, we can use time to calculate distance traveled or restaurant rating to calculate profit. Because we’re free to plug in whatever value we want for time of rating, we call these quantities variables. Furthermore, we use the term function to describe relationships like these – that is, ones that associate one value to each value the input variable could have. As you will see, the input doesn’t have to be just a variable but can be an expression, so we more generally refer to a function’s input as its argument

If the function is f and its input is x, we write the function as f(x) to make it clear that the value of f depends on the value of x. For example, if we wanted f to be a function that doubles and adds one, we would write:

f(x) = 2x + 1

We could just as well write:

f(a) = 2a + 1

The variable in the argument is just a placeholder. In fact, we could even insert another expression in its place, such as:

f(b − 3) = 2(b − 3) + 1

The notation just means, “When you replace the placeholder variable in the parentheses with something, replace it with the same something on the other side of the equation.”

Rules for Exponents

For any numbers a, b and c, these rules apply:

1.  ab * ac  = a(b+ c) 

We define it this way because this is what we’d expect for positive integers:

a2  * a3  = (a * a) * (a * a * a) = (a * a * a * a * a) = a5 

Now what should a negative exponent mean? Using the first rule:

a = a(2 - 1)  = (a * a) * a-1 

Divide both sides by a and we have:

1 = a * a − 1

Divide both sides by a again to find:

a-1  = 1/a

We can raise this to any power, b, and use the first rule again to find the second rule:

2. a-b  = 1/ab 

That also shows us what we get when we raise to the power of zero:

a0  = a(1 - 1) = a * a-1  = a/a = 1

3. a0  = 1

Now we know how exponents work for any integer. To go further, we need to think about integers again. Let’s raise a number to two powers:

(a3)2  = (a * a * a) * (a * a * a) = a(3 + 3)  = a(3 * 2) 

We, therefore, write the next rule:

4. (ab)c = a(b * c) 

What about fractional exponents? We use the rule we just found:

a = a1 = a(b/b) = (ab)(1/b)

So raising to a fractional power “undoes” raising to a whole number power. For example: a^(1/2) is the square root of a.

Be careful, though, squaring is not invertible:

a2 = (-1)2 * a2 = (-a)2

So, in general, square roots (and any other even-number roots) can be either positive or negative. Keeping that in mind:

5. (a)(b/c) = c√ab

c√ab means ”the cth root of ab”, and we call the “√” symbol for the cth root a radical.

One more rule, thinking of positive numbers one more time:

(a * b)2 = (a * b) * (a * b) = (a * a)(b * b) = (a2)(b2)

Making this a general rule, we have

6. (a * b)c  =(ac)(bc)

Exponential Functions

Anytime a rate of some quantity’s change is proportional to the quantity itself, an exponential function describes the situation. The general formula is:

(current amount) = (initial amount)(proportion of change)time

For example, if I earn 5% interest on an investment every year and I start out by investing $200, the formula for how much the investment will have grown to in t years is:

200 * (1 + 0.05)t

Note that the proportion of the change is 1.05, not 0.05. The 1 accounts for the amount already in the investment and the 0.05 accounts for the interest.

We can have a different variable in the place of time in the formula above. As another example, suppose you earn twice as many points in a bean bag toss game for every ten feet further back you are when throwing, starting with 10 points at the closest allowed distance. Then the number of points you could earn with a throw x feet behind the closest line is:

10 * 2(x/10)

Radical Equations

If we’re faced with an equation involving a radical, we need to get rid of it by raising each side to the power corresponding to the root in the radical. For example, if we have a square root in an equation, we need to square that square root to get out whatever is inside it. Let’s solve for x in this equation:

3 = 1 + √(x − 5)

If we square it right now, when we multiply out everything on the right side we’ll have: 1 + 2√(x − 5)) + (x − 5) and are still stuck with the radical. We need to isolate the radical first. Subtract 1 from each side and we get:

2 = (x − 5)

Now we can get rid of the root by squaring both sides:

4 = x − 5

x = 9

We can plug that back in and check that the equation holds. It’s a good idea to check because, as we mentioned earlier, we can’t just “undo” squaring. There are both positive and negative roots.

Polynomials

Polynomials are functions that can be written as sums of powers of variables. For example, these are all polynomials:

x2  + x − 2

x5  + x4  − 7x3  + x − 4

x + 1

2 = 2x0  

The powers in polynomials could be any fixed numbers, but not variables, and polynomials can have any number of terms. In general, we can write a polynomial like this:

anxn + an - 1xn - 1 + … a2x2 + a1x + a0

where n is any positive integer and each of a1, a2, and all the rest up to an is any constant number (possibly zero). This isn’t the only way to write polynomials, though, as we will soon see.

Quadratic Equations and Functions

Quadratic equations are a specific type of polynomial equation that only go up to the second power of their variable, so the most general form is ax2 + bx + c, where a, b, and c are any constant numbers. We give them a special name because they appear more frequently and are easier to analyze than polynomials in general.

Methods of Solving

Suppose we’re given a quadratic equation as ax2 + bx + c = 0 and asked to solve it. There are several different ways to think about this problem. If you can identify the factors, that’s the fastest method. First, divide through by a. If we can factor the quadratic as:

x2 + (b/a)x + c/a = (x + s)(x + t)

that can be expanded to:

x2 + (b/a)x + c/a = x2 + (s + t)x + st

If you can think of two numbers s and t, for which the two sides of the equation match, then the solutions are x = −s and x = −t, since, for either of those, (x + s)(x + t) = 0.

We can’t always factor quadratic equations with real numbers, though, and even when we can they might be difficult to see, so it would be wise to think of a more general approach as a backup. Fortunately, we have another trick to use on quadratic equations known as completing the square, which goes as follows.

We can easily calculate that

(x + r)2 = x2 + 2rx + r2

For this to look like x2 + (b/a)x + c/a, r must be b/2a

Then, for the last term, we subtract: r2 = b2/4a2 and add c/a, so:

(x + r)2 − r2 + c/a = x2 + 2rx + c/a = x2 + (b/a)x + c/a

as we wanted. Now we can use ax2 + bx + c  = 0 to solve for x. Dividing by a and completing the square with r = b/2a, we have:

x + b2/2a − b2/4a2 + c/a = 0

Isolating the square and taking the square root of both sides:

x + b/2a = ±√(b2/4a2 − c/a)

Remember that we have to consider both the positive and negative root. Finally, subtract b/2a from both sides and simplify:

x = −b/2a ± √(b2/4a2 − c/a) = (−b ± √(b2 − 4ac))/2a

This formula is known as the quadratic formula, and we just derived it by completing the square. It works for any quadratic equation of the form ax2 + bx + c = 0 as long as b^2 ≥ 4ac

If b2 < 4ac, then the equation has no real solutions.

What if b2 = 4ac? In that case, the square root part is zero and the two solutions merge to the same value.

If you’re having trouble with a quadratic or any other function, plugging in numbers and sketching graphs are often helpful. Doing so can help you to get an idea of how functions behave.

Creating a Function

If you’re given some information and asked to solve for an unknown quantity, the procedure is the same as it always has been: substitute values into the equation that relates them. For example, suppose you want to make a rectangular castle with an area of 40 square feet inside. Its thick stone walls take up some space, so if the castle measures 2x by x feet on the outside, the open space inside has length 2x−2 and width x−2. What will the dimensions of the castle be (as measured from the outside)?

Area = length * width, so 40 = (2x − 2)(x − 2) = 2x2 − 6x + 4

Subtract 40 and divide by 2 on each side and we have:

x2 − 3x − 18 = 0

We can factor this to (x − 6)(x + 3) = 0 or use the quadratic formula to find the possible solutions, x = 6 and x = −3. Negative length and width make no sense, so we need x = 6, which means the castle will be 12 feet by 6 feet.

Rational Equations and Linear Expressions

Rational functions are formed by dividing one polynomial by another. Like polynomials, they can be written in more than one way. Remember when we multiplied two polynomials and used the rules of exponents to find that their product was another polynomial, whose highest exponent was the sum of the two exponents we started with? When we perform a process known as polynomial long division on a rational function, we essentially are trying to reverse this process.

The rational function’s terms, then, have exponents no higher than the largest in its numerator minus the largest in its denominator. In long division with numbers, we could be left with a remainder that was too small to be divided by the divisor. In polynomial division, the remainder is a rational function in which the numerator’s highest power is lower than the denominator’s highest power.

Dividing Polynomials by Linear Expressions

Linear equations are just polynomials that don’t go any higher than the first power of x, so a polynomial divided by a linear expression is an example of a rational function. Some questions in the Passport to Advanced Math section may ask you to expand one of these using polynomial long division, which is similar in spirit to numerical long division. You can use the similarity to help yourself remember the procedure for polynomial division, and it’s easier to see if we use examples.

Doing long division for 635/4, we see that we can subtract 6 − 4 = 2; the six is in the hundreds place so the 4 should be also. What we’re really doing is shorthand for subtracting 600 − 4 * 100 = 200.

Next, we take the tens place digit and, since 4 goes into 23 five times, we write 5 in the tens place. Again, what we’re really doing is 230 − 4 * 50 = 30.

Finally, we take the 5 down to make 35 and write an 8 in the ones place: 35 − 4 * 8 = 3, leaving a remainder of 3.

What we did at each step was get rid of the number in the highest place we could by multiplying the divisor by the appropriate factor. This idea extends to polynomial division as well. For example, let’s expand this:

(4x2 + 4x + 3)/(2x − 1)

Begin by considering what must be multiplied with 2x − 1 to match the first term of 4x2 in the numerator: 2x * 2x = 4x2, so the first term in the quotient will be 2x. Multiply the divisor by this 2x and subtract this value from the numerator:

(4x2  + 4x + 3) − (4x2 + 2x), which gives 6x + 3.

The denominator must be multiplied by 3 to divide into 6x. Multiplying the denominator by 3 gives 6x − 3, and because we are subtracting this, the signs switch to give −6x + 3, which when combined gives a remainder of 6 which is then placed in a fraction as the numerator with the original divisor as the denominator to produce the final quotient and remainder:

(4x2 + 4x + 3)/(2x − 1) = (2x + 3 + 6)/(2x−1)

Variables in the Denominator

When faced with an equation with variables in the denominators of some terms, the most helpful approach is generally to multiply by the denominator to get rid of them and leave everything in terms of polynomials, which we can then solve with the techniques we discussed earlier. For example, for an equation like this:

5/(b + 1) = (3 − 1)/(b + 2)

Multiply by (b+1)(b+2) and the equation becomes:

5(b + 2) = (3 − 1)(b + 1)

5b + 10 = 3b + 3 − b − 1

5b + 10 = 2b + 2

And solving for b:

3b = −8

b = −8/3

Systems of Equations

When we have multiple equations relating the same variables, we call the collection of equations a system of equations. Solving for the variables means we must find values for each variable for which every equation in the system holds true.

Equations in Two Variables

Each equation relating some variables places an additional constraint on the variables. For example, if y = x + 2, then as soon as we know what x is we know what y must be. Except in special cases when some equations mean the same thing (like 2x + 6 = 4 and x + 3 = 2), we have exactly one solution when the number of variables and the number of equations is the same.

In the Heart of Algebra study guide, we cover how to solve a system of two linear equations by substitution or subtraction. The same ideas apply to more general equations. Try to isolate one of the variables so you can solve for its value. If you can get isolate a variable, substitution works the same way as before. For example, to solve the system of:

x + y = 2

and

(y − 1)2 + xy = 1

we can isolate x = 2 − y and substitute this into the other:

1 = (y − 1)2 + xy = (y − 1)2 + (2 − y)y = y2 − 2 + 1 + 2 y − y2 = 2y − 1

Now we just have 

2y − 1 = 1, so y = 1.

Then, x = 2 − y = 1, so the solution is (x, y) = (1, 1).

As an example in which we might rather use the form of the equations rather than isolating a variable, consider this system:

(x − 8)2 − x = 5y + 2

and

x = 2(y + 4)

We could expand (x − 8)^2, but it would be faster to notice that x = 2(y + 4) can be rewritten as x − 8 = 2y. Then we can insert that into the other equation to find:

5y + 2 = (x − 8)2 − x = (2y)2 − (2y + 8) = 4y2 − 2y − 8

so

0 = 4y2 − 7y − 6

We still need to solve a quadratic equation for y and then use the linear equation to find x, but this way, we take a shortcut by not having to expand the square.

Function Representation Relationships

In some of these questions, you will rely heavily on the ability to coordinate graphic and algebraic representations of the same function. Be sure you understand the roles of these components in this type of math.

Graph of a Function

As we said earlier, a function takes input values and associates them with output values. A useful notation for this is (x, f(x)) where x is the input and f(x) is the output. It’s often helpful to visualize functions, so we draw graphs by associating x with the horizontal coordinate and f(x) with the vertical coordinate. In general, we call the horizontal coordinate x and the vertical coordinate y, and the function is drawn by marking all points (x, y) at which f(x) = y.

Intercepts

An intercept is the place where a function encounters an axis. We usually are interested in x- and y-intercepts, i.e. where the function crosses the x- or y-axis.

The intercepts are determined by only one coordinate since the other must be zero — the y-axis is the line where x = 0, so a y-intercept of 3 means that the function crosses the y-axis at the point (0, 3). Similarly, the x-intercept is the line where y = 0.

The function graphed in blue here has x-intercepts at (−2, 0), (1, 0), and (3, 0) and a y-intercept at (0, 6), indicated by the red lines.

Range and Domain

The domain of a function is the set of values it can take as an argument, while a function’s range is the set of all values the function can give as output. Their names may seem strange at first, but they actually make sense with the right analogy. The function’s domain is where the function can be used, like a king only having power over his domain, and the range is where the function can reach, like how far his archers can shoot from inside their king’s domain. On a graph, the domain would be the set of all x-coordinates that are points on the graph and the range is the set of all y-coordinates that are points on the graph.

Increasing and Decreasing

Maximum and minimum values are exactly what their names say: the highest and lowest values a function attains, respectively. On a graph, they’re easy to identify—the maximum is the highest a function goes, and the minimum is the lowest it goes. Just be careful not to confuse the function’s maximum value, which is the y-coordinate of the highest point, with where it reaches the maximum, which is the x-coordinate. The same goes for the minimum. Of course, functions may not have one or either of them. Many functions just keep increasing or decreasing.

End Behavior

End behavior refers to how a function behaves as its argument increases or decreases as far as we can take it. In a graph, this is what the function does toward either end, looking from left to right. For example, the function below approaches infinity as its argument grows larger, and it approaches zero as the argument becomes more negative.

Asymptotes

Asymptotes are lines that functions approach but never quite touch. In the example above, the line y = 0 is an asymptote. More generally, the asymptote is y = b, where b is the value y approaches. Asymptotes can be horizontal, vertical, or even slanted. Below is a function in blue with a slanted asymptote in red. It also has a vertical asymptote at x = 0. A rational function’s vertical asymptotes can be found by looking for where its denominator is zero. As the denominator becomes smaller, the value of the fraction itself becomes larger.

Symmetry

Sometimes a function will look like a mirror image. In mathematical terms, we call a function symmetric about an axis if its value at any distance to one side of the axis is the same as the value at the same distance to the other side.

In function notation, f is symmetric with respect to the axis x = a if f(x − a) = f(−(x − a)).

Transformations

Suppose we have a function f(x). Since the horizontal axis is associated with a function’s input x, we can move everything by shifting the input: changing f(x) to f(x + a) moves the graph horizontally by units. Likewise, since the vertical axis is associated with the output f(x), we can shift the function vertically by changing f(x) to f(x) + a.

Be careful with the direction of the shift as the graph will move in the direction opposite to the sign of a.

For example, the value of f(x − 2) when x = 3 is the same as the value of f(x) when x = 1, so a negative number means a positive shift. After all, x has to increase to “balance out” the negative shift from −2. The same logic applies to vertical shifts as well.

Using Functions with Context

One guiding principle of the SAT exam is to relate questions to real-world situations and the math section is no exception. This becomes rather complex when you are dealing with things like functions and complicated equations, but you need to become fluent in doing so.

Example 1

The distance d traveled by an object starting with speed v with constant acceleration over a period of time t is given by d = (vt + at2)/2. Using this information, what is the expression for how much time it would take for an object starting from a standstill to travel a known distance d with acceleration a?

Here we are asked to solve for t in terms of the other variables. We’re told to assume we know the values of d and a and, since the object is starting from a standstill, its initial speed is v = 0. Then the equation is d = (at2)/2 and we can solve for t by multiplying by 2, dividing by a, and taking the square root: t = √(2d/a).

Example 2

The maximum range a projectile can be sent for a given initial speed is reached when launched at a 45° angle and is given by d = v2/g where d is range, v is initial speed and g is acceleration due to gravity. If the energy stored in an elastic band is completely transformed into kinetic energy, the distance the band is stretched is related to the projectile’s speed by (kx2)/2 = Energy = (mv2)/2, where k is the spring constant, x is the distance stretched and m is the projectile’s mass. How would the spring constant need to be changed if you wanted to have the same range while pulling only half as far back on a slingshot using the same projectile?

First, we need to identify which variables we know and which we don’t. We’re asked how the spring k constant would change, so that’s the variable we’d like to solve for. On the other hand, using the same projectile and range means that we can treat the mass m and range d as known. We also know the distance stretched, since that’s just the original distance, divided by 2, and that gravity is essentially the same unless space travel is involved.

Let’s start by solving for the spring constant. We find k=(mv2)/(x2). We know m and x but not v, so we need to check what we know about v from the other variables. In the formula d = (v2)/g, neither of the other variables (d and g) change, so v must not either. Now we’re ready to answer the question: When everything else remains the same, how does k change as a function of x? Specifically, if we replace x with x2, what happens to k? We can readily calculate this.

Calling the new spring constant k_new:

k_new = (mv2)/(x/2)2 = (mv2)/((x2)/4) = (4mv2)/(x2)

Now mv2x2 was the original spring constant, so knew is four times the original spring constant. The answer, then, is that the spring constant must quadruple, or change by a factor of four.

We could also have solved for v2 and substituted this into the preceding expression to obtain an explicit formula for d in terms of the variables we’re told not to change (k, x, m, and g). Then we would have found the relationship d = kx2/mg, which works just as well and is simply another perspective for the problem, which may help to make it more clear.