PRE CALCULUS Study Guide 2026

Everything you need to pass the PRE CALCULUS exam in one place: the exam format, every topic to study, real practice questions with explanations, flashcards, and full-length practice tests. Free, no sign-up needed.

📋 PRE CALCULUS Exam Format at a Glance

48
Questions
90 min
Time Limit
65.00%
Passing Score

📚 PRE CALCULUS Topics to Study (21)

✍️ Sample PRE CALCULUS Questions & Answers

1. What distinguishes an ellipse from a hyperbola in standard form?
Ellipse uses + between terms; hyperbola uses −

The standard ellipse equation has the form x²/a² + y²/b² = 1, while the hyperbola uses a minus sign.

2. What is the standard form of a circle with center (h, k) and radius r?
(x−h)² + (y−k)² = r²

The standard equation of a circle is (x−h)² + (y−k)² = r², where (h, k) is the center.

3. The diagram that follows has been color-coded. In this diagram, the different circle parts are designated with the letters From A to F. Which circle part corresponds to the letter C?
Radius

A radius is a line segment that extends from the exact center of a circle to any point on its circumference. It represents half the length of the diameter. If letter C in the diagram points to such a segment, it is the radius.

4. For the hyperbola x²/16 − y²/9 = 1, what are the equations of the asymptotes?
y = ±(3/4)x

The asymptotes of x²/a² − y²/b² = 1 are y = ±(b/a)x, so y = ±(3/4)x.

5. What is -27's cube root?
-3

The cube root of a number 'x' is a value 'y' such that y multiplied by itself three times equals x (y³ = x). For -27, we need a number that, when cubed, results in -27. Since (-3) * (-3) * (-3) = 9 * (-3) = -27, the cube root of -27 is -3.

6. f(x) = x3 - 9x^2 - 45x - 27 : Determine the remaining zeros if -3 is a zero of f(x).
{-3, 6 ± 3√5}

Since -3 is a zero of f(x) = x^3 - 9x^2 - 45x - 27, (x+3) is a factor. Using synthetic division with -3, we divide the polynomial to get the quotient x^2 - 12x - 9. To find the remaining zeros, set this quadratic equal to zero and use the quadratic formula: x = [12 ± √((-12)^2 - 4(1)(-9))] / 2(1) = [12 ± √(144 + 36)] / 2 = [12 ± √180] / 2. Simplifying √180 to 6√5, we get x = [12 ± 6√5] / 2 = 6 ± 3√5. Therefore, the remaining zeros are 6 + 3√5 and 6 - 3√5.

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