SAT Math Concise Study Notes : Algebra and Functions

SAT MAth and English  – full syllabus practice tests

  • Algebra Weightage: 35%  Questions: 13-15
  • Advanced Math Weightage: 35% Questions: 13-15
  • Problem-solving and Data Analysis Weightage: 15%  Questions: 5-7
  • Geometry and Trigonometry Weightage: 15% Questions: 5-7

Polynomials are algebraic expressions involving numbers and variables. You should already know how to factor and multiply polynomials. Keep in mind the difference of squares formula:

$
a^2-b^2=(a+b)(a-b)
$

  •  If \(x^2-y^2=25.6\) and \(x+y=8\), what is the value of \(2 x-2 y\) ?

$
\begin{gathered}
x^2-y^2=(x+y)(x-y)=25.6 \\
8(x-y)=25.6 \\
x-y=3.2 \\
2 x-2 y=2(x-y)=2 \times 3.2=6.4
\end{gathered}
$

Remember what the question is asking-for this problem, it would have been a waste of time to solve for \(x\) and \(y\) individually!

You may need to solve single equations and systems of equations. To solve a system of equations, you can isolate a variable in one equation and substitute the resulting expression into the other equation. Sometimes you can also save time by adding or subtracting two or more equations.

  •  If 4 times a number minus a second number is equal to 1 , and the sum of the two numbers is 9 , what is their product?

Create a system of equations:
$
\begin{gathered}
4 x-y=1 \\
x+y=9
\end{gathered}
$

Add:

$
5 x=10
$

Solve:

$
\begin{aligned}
x & =2 \\
y & =9-2=7 \\
x y & =2 \times 7=14
\end{aligned}
$

Work with inequalities and systems of inequalities the same way you would work with equations and systems of equations, but keep in mind several key properties. Any operation involving positive numbers has no effect on the inequality, but multiplying or dividing by a negative number reverses the inequality. If both sides of an inequality are positive or negative, taking their reciprocal also reverses the inequality.
$
\text { If } a<b \text {, then }-a>-b
$

If \(a<b\) and \(c\) is negative, then \(a c>b c\) and \(\frac{a}{c}>\frac{b}{c}\)

If \(a<b\) and both \(a\) and \(b\) are the same sign, then \(\frac{1}{a}>\frac{1}{b}\)

  • If \(x^2+y<5\) and \(y>3\), give one possible value for \(x\).

$
\begin{gathered}
x^2+y<5 \text {, so } y<5-x^2 \\
y>3 \text {, so } 3<y<5-x^2 \\
3<5-x^2 \\
-2<-x^2 \\
2>x^2 \\
-\sqrt{2}<x<\sqrt{2} \\
x=1, \frac{1}{2},-\frac{1}{2},-1 \ldots
\end{gathered}
$

You will frequently need to convert word problems into algebraic equations. Look for common words and phrases that correspond to mathematical operations:

  •  The total of Jake’s and Amy’s ages is 17. Last year, Amy was twice Jake’s age. How old is Jake now?

Key words: total, last year, and twice
Let \(j\) represent Jake’s current age and \(a\) represent Amy’s current age:
$
j+a=17
$

To represent the relationship between their ages last year, subtract 1 from each of their current ages:
$
a-1=2(j-1)
$

Solve for \(j\) :
$
\begin{gathered}
a=17-j \\
17-j-1=2(j-1) \\
16-j=2 j-2 \\
18=3 j \\
j=6
\end{gathered}
$

Jake is currently 6 years old.

  •  An event has two admission prices: adult tickets are \(\$ 10\) each and student/senior tickets are \(\$ 8\) each. The ticket office has sold 77 tickets totalling \(\$ 686\). How many

adult tickets were sold?
Key words: each, totalling
Let \(a\) represent the number of adult tickets and \(s\) represent the number of student/senior tickets:
$
a+s=77
$

To represent total revenue, multiply each type of ticket by its price:
$
10 a+8 s=686
$

Solve for \(a\) :
$
\begin{gathered}
s=77-a \\
10 a+8(77-a)=686 \\
10 a+616-8 a=686 \\
2 a=70 \\
a=35
\end{gathered}
$

There were 35 adult tickets sold.

A function is a rule that associates one set of numbers with another dependent set of numbers. The function \(f(x)=x+1\) tells us that every real number \(x\) is assigned to the corresponding number \(x+1\). If you are asked to find \(f(3)\), simply follow the rule connecting 3 to its assigned number: \(f(3)=3+1=4\)

  •  If \((x)=3 x^2-2 x\), what is \(f\left(\frac{1}{2}\right)+f(1)\) ?

$
\begin{aligned}
& f\left(\frac{1}{2}\right)=3\left(\frac{1}{2}\right)^2-2\left(\frac{1}{2}\right)=\frac{3}{4}-1=-\frac{1}{4} \\
& f(1)=3(1)^2-2(1)=3-2=1 \\
& f\left(\frac{1}{2}\right)+f(1)=-\frac{1}{4}+1=\frac{3}{4}
\end{aligned}
$

  •  If \(f(x)=3 x^2-2 x\), what is \(f\left(x^2\right)\) ?

$
f\left(x^2\right)=3\left(x^2\right)^2-2\left(x^2\right)=3 x^4-2 x^2
$

  • If \(f(x)=3 x^2-2 x\), what is \(f(f(2))\) ?

$
\begin{aligned}
& f(2)=3(2)^2-2(2)=12-4=8 \\
& f(8)=3(8)^2-2(8)=192-16=176 \\
& f(f(2))=176
\end{aligned}
$

The domain of a function is the set of all the numbers for which the function is defined-that is, all the “input” numbers for which the function still works. The range of a function is the set of all the “output” numbers.

  • What is the domain of the function \((x)=(x+2)^{\frac{1}{2}}\) ?

$
f(x)=(x+2)^{\frac{1}{2}}=\sqrt{x+2}
$

Because we can’t take the square root of a negative number, we know that \(x+2\) can’t be negative:
$
\begin{gathered}
x+2 \geq 0 \\
x \geq-2
\end{gathered}
$

  • What number is not in the domain of the function \(f(x)=\frac{1}{x^2}\) ?

We can take the square of any real number \(x\), but we can’t divide by zero. Therefore:

$
\begin{aligned}
x^2 & \neq 0 \\
x & \neq 0
\end{aligned}
$

Zero is the only number not in the domain of our function.

Instead of using function notation to represent the relationships between sets of numbers, the SAT will often use strange symbols: smiley faces, stars, bubbles, etc. Treat these the same way you would treat function questions—look for the rule, and figure out how you should apply it.

  •  For all negative values \(x\), let \(\diamond x=4 x+7\). For zero and all positive values \(x\), let \(\diamond x=3 x^2\). What is the value of \(\diamond \diamond(-1)\) ?

$
\begin{aligned}
& \diamond(-1)=4(-1)+7=3 \\
& \diamond(3)=3\left(3^2\right)=27 \\
& \diamond \diamond(-1)=27
\end{aligned}
$

  •  \(\quad\) For all non-zero numbers \(a\) and \(b\), let \(\Delta\) be defined as

$
a \Delta b=\frac{2 a}{b}
$

If \(3 \Delta 4=x \Delta 2\), what is the value of \(x\) ?
$
\begin{aligned}
3 \Delta 4 & =\frac{2 \times 3}{4}=\frac{6}{4} \\
x \Delta 2 & =\frac{2 x}{2}=x \\
\frac{6}{4} & =x \\
x & =\frac{3}{2}
\end{aligned}
$

  • Let  represent the number of factors of any real number x. For example, 6 = 4 because
    6 has four factors: 1, 2, 3 and 6. What is 9?

The graph of a function is the collection of all ordered pairs \((x, y)\) where \(y=f(x)\). Thus, the graph of \(f(x)=x+1\) is the line \(y=x+1\). You will never be asked to draw a graph on the SAT, but you may need to derive an equation or function based on a graph.

Linear functions can be represented in the form \(y=m x+b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. Remember the formula for calculating slope:
$
m=\frac{\text { rise }}{\text { run }}=\frac{y_2-y_1}{x_2-x_1}
$

Facts about slope:

  • Horizontal lines have slopes of zero.
  • Vertical lines have undefined slopes.
  •  Non-vertical parallel lines have equal slopes.
  •  Non-vertical perpendicular lines have slopes whose product is -1 .
  • A line with a slope of 5 passes through the points \((k, 7)\) and \((-1,-3)\). What is the value of \(k\) ?

$
\begin{gathered}
5=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-7}{-1-k} \\
5(-1-k)=-10 \\
-1-k=-2 \\
k=1
\end{gathered}
$

  • In the \(x y\)-coordinate plane, line \(h\) has a slope of 2 and is perpendicular to line \(g\). If line \(g\) passes through the origin and intersects line \(h\) at the point \((a, 3+a)\), what is the value of \(a\) ?

Let \(m\) represent the slope of line \(g\). Line \(g\) and line \(h\) are perpendicular, so:
$
\begin{gathered}
2 m=-1 \\
m=-\frac{1}{2}
\end{gathered}
$

We know that line \(g\) passes through the origin, so its \(y\)-intercept is 0 . Its equation must be:
$
y=-\frac{1}{2} x
$

Plug the point \((a, 3+a)\) into this equation to solve for \(a\) :
$
\begin{gathered}
3+a=-\frac{1}{2} a \\
\frac{3}{2} a=-3 \\
a=-2
\end{gathered}
$

Quadratic functions can be represented in the form \(y=a x^2+b x+c\), where \(c\) is the \(y\)-intercept. The sign of \(a\) determines whether the parabola opens upwards ( \(a\) is positive) or downwards ( \(a\) is

negative). You can often find the \(x\)-intercepts of a parabola by factoring; the SAT will never require you to use the quadratic formula.

The vertex of a parabola is its lowest or highest point. If we call the coordinates of the vertex \((h, k)\), then:
$
h=\frac{-b}{2 a}
$

When asked to solve for other points on a parabola, remember that a parabola is symmetrical about the vertical line through its vertex.

  •  In the figure below, \(f(x)=x^2-1\) and \(g(x)=-x^2+c\), where \(c\) is a constant. The two functions intersect at points \(A\) and \(B\). If \(\overline{A B}=4\), what is the value of \(c\) ?

Because a parabola is symmetrical and \(\overline{A B}=4\), each point of intersection must be 2 units away from the \(y\)-axis. Thus, we can call their coordinates \((-2, y)\) and \((2, y)\). Solve for \(y\) by plugging one of these points into the first equation:
$
\begin{gathered}
f(x)=x^2-1 \\
y=2^2-1 \\
y=3
\end{gathered}
$

The two equations intersect at \((-2,3)\) and \((2,3)\). Plug one of these points into the second equation to solve for \(c\) :
$
\begin{gathered}
g(x)=-x^2+c \\
3=-2^2+c \\
c=7
\end{gathered}
$

You may be asked to analyze the behaviour of graphs without knowing or using their equations.

  • The graph of \(y=f(x)\) is shown below. If \(f(2)=d\), what is \(f(d)\) ?

$
\begin{gathered}
f(2)=4, \text { so } d=4 \\
f(d)=f(4)=0
\end{gathered}
$

A translation of a graph is a rigid shift vertically or horizontally that does not change the size or shape of the graph. For the graph of \(y=f(x)\), a vertical translation of \(a\) units and a horizontal translation of \(b\) units results in the equation
$
y-a=f(x-b)
$

Let’s begin with the graph of \(y=x^2\). If we shift vertically 3 units and horizontally 2 units, we get \(y-3=(x-2)^2\). If we shift vertically -3 units and horizontally -2 units, we get \(y+3=(x+2)^2\).

  •  Below are the graphs of the functions \(f(x)=x^3-2 x\) and \(g(x)=f(x-c)+d\), where \(c\) and \(d\) are both constants. What is the value of \(d+c\) ?

In the equation \(y-a=f(x-b)\), \(a\) represents the vertical shift and \(b\) represents the horizontal shift. Re-write the equation of \(g(x)\) to look similar:
$
g(x)-d=f(x-c)
$

Here, \(d\) represents the vertical shift and \(c\) represents the horizontal shift. By comparing the two points given on the graph, we can see that the graph has been translated 2 units horizontally and 4 units vertically. Thus:
$
\begin{gathered}
c=2 \\
d=4 \\
d+c=6
\end{gathered}
$

 

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