## SAT SUBJECT TEST MATH LEVEL 2

## PART 2

## REVIEW OF MAJOR TOPICS

## CHAPTER 1

Functions

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1.2 Polynomial Functions

### 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 3*x* + *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)** 2*x*^{2 }+ *x *– 6 = (2*x *– 3)(*x *+ 2) = 0. The zeros are and –2.

5. **(D)** Sum of .

6. **(E)** From the discriminant *b*^{2 }– 4*ac *= 4 – 4 · 1 · 3 = –8 < 0.

* 7. **(B)** The equation of a vertical parabola with its vertex at the origin has the form *y *= *ax*^{2 }. Substitute (7,7) for *x *and *y *to find . When *y *= 6, *x*^{2 }= 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) = 1^{3} – 3 · 1^{2} + 2 · 1 – 4*b *= 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) 3*x*^{2} – *x* – 2 = (3*x* + 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)** *x*^{2 }– 16*x *+ 48 = (*x *– 4)(*x *– 12) = 0, when *x *= 4 or 12. Numbers between these satisfy the original inequality.