Barron's SAT Subject Test Math Level 2, 10th Edition (2012)

Part 2. REVIEW OF MAJOR TOPICS

Chapter 1. Functions

1.2 Polynomial Functions

LINEAR FUNCTIONS

Linear functions are polynomials in which the largest exponent is 1. The graph is always a straight line. Although the general form of the equation is Ax + By + C = 0, where A, B, and C are constants, the most useful form occurs when the equation is solved for y. This is known as the slope-intercept form and is written y = mx + b. The slope of the line is represented by m and is defined to be the ratio of , where (x₁, y₁) and (x₂, y₂) are any two points on the line. The y-intercept is b (the point where the graph crosses the y-axis).

If you solve the general equation of a line, you will find that the slope is and the y-intercept is .

You can always quickly write an equation of a line when given its slope and a point on it by using the point-slope form: y – y₁ = m(x – x₁), where m is the slope and (x₁,y₁) is the point. If you are given two points on a line, you must first find the slope using the two points. Then use either point and this slope to write the equation. Once you have the equation in point-slope form, you can always solve for y to get the slope-intercept form if necessary.

EXAMPLES

1. Write an equation of the line containing (6,–5) and having slope .

In point-slope form, the equation is .

2. Write an equation of the line containing (1,–3) and (–4,–2).

First find the slope . Then use the point (1,–3) and this slope to write the point-slope equation .

Parallel lines have the same slope. The slopes of two perpendicular lines are negative reciprocals of one another.

3. The equation of line l₁ is y = 2x + 3, and the equation of line l₂ is y = 2x – 5.

These lines are parallel because the slope of each line is 2, and the y-intercepts are different.

4. The equation of line l₁ is , and the equation of line l₂ is .

These lines are perpendicular because the slope of l₂,, is the negative reciprocal of the slope of l₁,

You can use these facts to write an equation of a line that is parallel or perpendicular to a given line and that contains a given point.

5. Write an equation of the line containing (1,7) and parallel to the line 3x + 5y = 8.

The slope of the given line is . The point-slope equation of the line containing (1,7) is therefore .

6. Write an equation of the line containing (–3,2) and perpendicular to y = 4x – 5.

The slope of the given line is 4, so the slope of a line perpendicular to it is . The desired equation is .

The distance between two points P and Q whose coordinates are (x₁,y₁) and (x₂,y₂) is given by the formula

Distance =

and the midpoint, M, of the segment has coordinates .

7. Given point (2,–3) and point (–5,4), find the length of and the coordinates of the midpoint, M.

EXERCISES

1. The slope of the line through points A(3,–2) and B(–2,–3) is

  (A) −5

  (B) −

  (C)

  (D) 1

  (E) 5

2. The slope of line 8x + 12y + 5 = 0 is

  (A)

  (B)

  (C)

  (D) 2

  (E) 3

3. The slope of the line perpendicular to line 3x – 5y + 8 = 0 is

  (A)

  (B)

  (C)

  (D)

  (E) 3

4. The y-intercept of the line through the two points whose coordinates are (5,–2) and (1,3) is

  (A)

  (B)

  (C)

  (D) 7

  (E) 17

5. The equation of the perpendicular bisector of the segment joining the points whose coordinates are (1,4) and (–2,3) is

  (A) 3x – 2y + 5 = 0

  (B) x – 3y + 2 = 0

  (C) 3x + y – 2 = 0

  (D) x – 3y + 11 = 0

  (E) x + 3y – 10 = 0

6. The length of the segment joining the points with coordinates (–2,4) and (3,–5) is

  (A) 2.8

  (B) 3.7

  (C) 10.0

  (D) 10.3

  (E) none of these

7. The slope of the line parallel to the line whose equation is 2x + 3y = 8 is

  (A) –2

  (B)

  (C)

  (D)

  (E)

8. If the graph of is perpendicular to the graph of ax + 3y + 2 = 0, then a =

  (A) – 4.5

  (B) –2.22

  (C) –1.35

  (D) 0.45

  (E) 1.35

QUADRATIC FUNCTIONS

Quadratic functions are polynomials in which the largest exponent is 2. The graph is always a parabola. The general form of the equation is y = ax² + bx + c. If a > 0, the parabola opens up and has a minimum value. If a < 0, the parabola opens down and has a maximum value. The x-coordinate of the vertex of the parabola is equal to , and the axis of symmetry is the vertical line whose equation is .

To find the minimum (or maximum) value of the function, substitute for x to determine y.

Thus, in general the coordinates of the vertex are and the minimum (or maximum) value of the function is .

Unless specifically limited, the domain of a quadratic function is all real numbers, and the range is all values of y greater than or equal to the minimum value (or all values of y less than or equal to the maximum value) of the function.

The examples below provide algebraic underpinnings of how the orientation, vertex, axis of symmetry, and zeros are determined. You can, of course, use a graphing calculator to sketch a parabola and find its vertex and x-intercepts.

EXAMPLES

1. Determine the coordinates of the vertex and the equation of the axis of symmetry of y = 3x² + 2x – 5. Does the quadratic function have a minimum or maximum value? If so, what is it?

The equation of the axis of symmetry is

and the y-coordinate of the vertex is

The vertex is, therefore, at .

The function has a minimum value because a = 3 > 0. The minimum value is . The graph of y = 3x² + 2x – 5 is shown below.

The points where the graph crosses the x-axis are called the zeros of the function and occur when y = 0. To find the zeros of y = 3x² + 2x – 5, solve the quadratic equation 3x² + 2x – 5 = 0. By factoring, 3x² + 2x – 5 = (3x + 5)(x – 1) = 0. Thus, 3x + 5 = 0 or x – 1 = 0, which leads to or 1.

Every quadratic equation can be changed into the form ax² + bx + c = 0 (if it is not already in that form), which can be solved by completing the square. The solutions are , the general quadratic formula. Substitute a = 3, b = 2, and c = –5 to get the same zeros, or x = 1.

2. Find the zeros of y = 2x² + 3x – 4.

Solve the equation 2x² + 3x – 4 = 0. The left side does not factor. Using the quadratic formula with a = 2, b = 3, and c = – 4, gives x 0.85078 or –2.35078. These solutions are most readily obtained by using the polynomial solver on your graphing calculator.

Note that the sum of the two zeros, equals , and their product equals . This information can be used to check whether the correct zeros have been found. In Example 2, the sum and product of the zeros can be determined by inspection from the equations. and Product .

At times it is necessary to determine only the nature of the roots of a quadratic equation, not the roots themselves. Because b² – 4ac of the general quadratic formula is under the radical, its sign determines whether the roots are real or imaginary. The quantity b² – 4ac is called the discriminant of a quadratic equation.

(i) If b² – 4ac = 0, the two roots are the same , and the graph of the function is tangent to the x-axis.

(ii) If b² – 4ac < 0, there is a negative number under the radical, which gives two complex numbers (of the form p + qi and p – qi, where ) as roots, and the graph of the function does not intersect the x-axis.

(iii) If b² – 4ac > 0, there is a positive number under the radical, which gives two different real roots, and the graph of the function intersects the x-axis at two points.

EXERCISES

1. The coordinates of the vertex of the parabola whose equation is y = 2x² + 4x – 5 are

  (A) (2, 11)

  (B) (–1, –7)

  (C) (1, 1)

  (D) (–2, –5)

  (E) (–4, 11)

2. The range of the function
f = {(x,y):y = 5 – 4x – x²} is

  (A) {y:y 0}

  (B) {y:y –9}

  (C) {y:y 9}

  (D) {y:y 0}

  (E) {y:y 1}

3. The equation of the axis of symmetry of the function y = 2x² + 3x – 6 is

  (A)

  (B)

  (C)

  (D)

  (E)

4. Find the zeros of y = 2x² + x – 6.

  (A) 3 and 2

  (B) –3 and 2

  (C) and

  (D) and 1

  (E) and –2

5. The sum of the zeros of y = 3x² – 6x – 4 is

  (A) –2

  (B)

  (C)

  (D) 2

  (E) 6

6. x² + 2x + 3 = 0 has

  (A) two real rational roots

  (B) two real irrational roots

  (C) two equal real roots

  (D) two equal rational roots

  (E) two complex conjugate roots

7. A parabola with a vertical axis has its vertex at the origin and passes through point (7,7). The parabola intersects line y = 6 at two points. The length of the segment joining these points is

 (A) 14

 (B) 13

 (C) 12

 (D) 8.6

  (E) 6.5

HIGHER-DEGREE POLYNOMIAL FUNCTIONS

Polynomial functions of degree greater than two (largest exponent greater than 2) are usually treated together since there are no simple formulas, such as the general quadratic formula, that aid in finding zeros.

Here are five facts about the graphs of polynomial functions:

1. They are always continuous curves. (The graph can be drawn without removing the pencil from the paper.)

2. If the largest exponent is an even number, both ends of the graph leave the coordinate system either at the top or at the bottom:

3. If the largest exponent is an odd number, the ends of the graph leave the coordinate system at opposite ends.

Facts (2) and (3) describe the end behavior of a polynomial.

4. If all the exponents are even numbers, the polynomial is an even function and therefore symmetric about the y-axis.

y = 3x⁴ + 2x² – 8

5. If all the exponents are odd numbers and there is no constant term, the polynomial is an odd function and therefore symmetric about the origin of the coordinate system.

y = 4x⁵ + 2x³ – 3x

A polynomial of degree n has n zeros. The zeros of polynomials with real coefficients can be real or imaginary numbers, but imaginary zeros must occur in pairs. For example, if the degree of a polynomial is 6, there are 6 real zeros and no imaginary zeros; 4 real zeros and 2 imaginary ones; 2 real zeros and 4 imaginary ones, or 6 imaginary zeros. Moreover, real zeros can occur more than once.

If a real zero occurs n times, it is said to have multiplicity n. If a zero of a polynomial has even multiplicity, its graph touches, but does not cross, the x-axis. For example, in the polynomial f(x) = (x – 3)⁴, 3 is a zero of multiplicity 4, and the graph of f(x) is tangent to the x-axis at x = 3. If a zero of a polynomial has odd multiplicity, its graph crosses the x-axis. For example, in the polynomial f(x) = (x – 1)⁵, 1 is a zero of multiplicity 5, and the graph of f(x) crosses the x-axis at x = 1.

The multiplicity of a real zero is counted toward the total number of zeros. For example, the polynomial f(x) = (x – 5)(x + 3)⁴(x² + 6) has degree 7: one zero (5) of multiplicity 1; one zero (–3) of multiplicity 4; and two imaginary zeros (i and –i ).

There are 5 facts that are useful when analyzing polynomial functions.

1. Remainder theorem—If a polynomial P(x) is divided by x – r (where r is any constant), then the remainder is P(r).

Divide P(x) = 3x⁵ – 4x⁴ – 15x² – 88x – 12 by x – 3.

  Enter P(x) into Y₁, return to the home screen, and evaluate Y₁(3) = –6. The remainder is –6 when you divide P(x) by x – 3.

2. Factor theorem —r is a zero of the polynomial P(x) if and only if x – r is a divisor of P(x).

Is x – 99 a factor of x⁴ – 100x³ + 97x² + 200x – 198?

  Call this polynomial P(x) and evaluate P(99) using your graphing calculator. Since P(99) = 0, the answer to the question is “Yes.”

3. Rational zero (root) theorem—If is a rational zero (reduced to lowest terms) of a polynomial P(x) with integral coefficients, then p is a factor of a₀ (the constant term) and q is a factor of a_n (the leading coefficient).

What are the possible rational zeros of P(x) = 3x³ + 2x² + 4x – 6?

  The divisors of the constant term –6 are ±1, ±2, ±3, ±6, and the divisors of the leading coefficient 3 are ±1, ±3. The 12 rational numbers that could possibly be zeros of P(x) are ±1, ±2, , the ratios of all numbers in the first group to all numbers in the second.

4. If P(x) is a polynomial with real coefficients, then complex zeros occur as conjugate pairs. (For example, if p + qi is a zero, then p – qi is also a zero.)

If 3 + 2i, 2, and 2 – 3i are all zeros of P(x) = 3x⁵ – 36x⁴ + 2x³ – 8x² + 9x – 338, what are the other zeros?

  Since P(x) must have five zeros because it is a fifth-degree polynomial, and since the coefficients are real and the complex zeros come in conjugate pairs, the two remaining zeros must be 3 – 2i and 2 + 3i.

5. Descartes’ rule of signs—The number of positive real zeros of a polynomial P(x) either is equal to the number of changes in the sign between terms or is less than that number by an even integer. The number of negative real zeros of P(x) either is equal to the number of changes of the sign between the terms of P(–x) or is less than that number by an even integer.

Three sign changes indicate there will be either one or three positive zeros of P(x).

One sign change indicates there will be exactly one negative zero of P(x).

EXERCISES

1. P(x) = ax⁴ + x³ – bx² – 4x + c. If P(x) increases without bound as x increases without bound, then, as x decreases without bound, P(x)

  (A) increases without bound

  (B) decreases without bound

  (C) approaches zero from above the x-axis

  (D) approaches zero from below the x-axis

  (E) cannot be determined

2. Which of the following is an odd function?

  I. f(x) = 3x³ + 5

  II. g(x) = 4x⁶ + 2x⁴ – 3x²

  III. h(x) = 7x⁵ – 8x³ + 12x

  (A) only I

  (B) only II

  (C) only III

  (D) only I and II

  (E) only I and III

3. How many possible rational roots are there for 2x⁴ + 4x³ – 6x² + 15x – 12 = 0?

  (A) 4

  (B) 6

  (C) 8

  (D) 12

  (E) 16

4. If both x – 1 and x – 2 are factors of x³ – 3x² + 2x – 4b, then b must be

  (A) 0

  (B) 1

  (C) 2

  (D) 3

  (E) 4

5. If 3x³ – 9x² + Kx – 12 is divisible by x – 3, then K =

  (A) –40

  (B) –3

  (C) 3

  (D) 4

  (E) 22

6. Write the equation of lowest degree with real coefficients if two of its roots are –1 and 1 + i.

  (A) x³ + x² + 2 = 0

  (B) x³ – x² – 2 = 0

  (C) x³ – x + 2 = 0

  (D) x³ – x² + 2 = 0

  (E) none of the above

INEQUALITIES

Given any algebraic expression f(x), there are exactly three situations that can exist:

 1. for some values of x, f(x) < 0;

 2. for some values of x, f(x) = 0;

 3. for some values of x, f(x) > 0.

If all three of these sets of numbers are indicated on a number line, the set of values that satisfy f(x) < 0 is always separated from the set of values that satisfy f(x) > 0 by the values of x that satisfy f(x) = 0.

EXAMPLE

Find the set of values for x that satisfies x² – 3x – 4 < 0.

Graph y = x² – 3x – 4. You need to find the x values of points on the graph that lie below the x-axis. First find the zeros: x = 4, x = –1. The points that lie below the x-axis are (strictly) between –1 and 4, or –1 < x < 4.

EXERCISES

1. Which of the following is equivalent to 3x² – x < 2?

  (A)

  (B)

  (C)

  (D)

  (E)

2. Solve x⁵ – 3x³ + 2x² – 3 > 0.

  (A)

  (B) (–1.90,–0.87)

  (C)

  (D) (–0.87,1.58)

  (E)

3. The number of integers that satisfy the inequality x² + 48 < 16x is

  (A) 0

  (B) 4

  (C) 7

  (D) an infinite number

  (E) none of the above

Answers and Explanations

Linear Functions

1. (C)

2. (B) . The slope is .

3. (A) . The slope of the given line is . The slope of a perpendicular line is .

4. (C) The slope of the line is , so the point-slope equation is . Solve for y to get . The y-intercept of the line is .

5. (C) The slope of the segment is . Therefore, the slope of a perpendicular line is –3. The midpoint of the segment is . Therefore, the point-slope equation is . In general form, this equation is 3x + y – 2 = 0.

* 6. (D) Length =

7. (C) . Therefore, the slope of a parallel line .

* 8. (C) The slope of the first line is , and the slope of the second line is . To be perpendicular, .

Quadratic Functions

1. (B) The x coordinate of the vertex is and the y coordinate is y = 2(–1)² + 4(–1) – 5 = –7. Hence the vertex is the point (–1,–7).

2. (C) Find the vertex: and y = 5 – 4(–2) – (–2)² = 9. Since a = –1 < 0 the parabola opens down, so the range is {y : y 9}.

3. (B) The x coordinate of the vertex is . Thus, the equation of the axis of symmetry is .

4. (E) 2x² + x – 6 = (2x – 3)(x + 2) = 0. The zeros are and –2.

5. (D) Sum of .

6. (E) From the discriminant b² – 4ac = 4 – 4 · 1 · 3 = –8 < 0.

* 7. (B) The equation of a vertical parabola with its vertex at the origin has the form y = ax² . Substitute (7,7) for x and y to find . When y = 6, x² = 42. Therefore, , and the segment = .

Higher-Degree Polynomial Functions

1. (A) Since the degree of the polynomial is an even number, both ends of the graph go off in the same direction. Since P(x) increases without bound as x increases, P(x) also increases without bound as x decreases.

2. (C) Since the exponents are all odd, and there is no constant term, III is the only odd function.

3. (E) Rational roots have the form , where p is a factor of 12 and q is a factor of 2.

  . The total is 16.

* 4. (A) Since x – 1 is a factor, P(1) = 1³ – 3 · 1² + 2 · 1 – 4b = 0. Therefore, b = 0.

* 5. (D) Substitute 3 for x set equal to zero and solve for K.

6. (D) 1 – i is also a root. To find the equation, multiply (x + 1)[x – (1 + i)][x – (1 – i)], which are the factors that produced the three roots.

Inequalities

1. (C) 3x² – x – 2 = (3x + 2)(x – 1) = 0 when or 1. Numbers between these satisfy the original inequality.

* 2. (C) Graph the function, and determine that the three zeros are –1.90, –0.87, and 1.58. The parts of the graph that are above the x-axis have x-coordinates between –1.90 and –0.87 and are larger than 1.58.

3. (C) x² – 16x + 48 = (x – 4)(x – 12) = 0, when x = 4 or 12. Numbers between these satisfy the original inequality.