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MB Pre-Calculus 40S study materials
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MB Pre-Calculus 40S practice questions

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Question 1Transformations

The graph of y = f(x) is translated 3 units to the left. Which equation gives the new graph?

  1. y = f(x) - 3
  2. y = f(x - 3)
  3. y = f(x) + 3
  4. y = f(x + 3) Correct answer.

Explanation

A horizontal translation replaces x with x - h for a shift of h units to the right, so a shift of 3 units left uses h = -3, giving f(x - (-3)) = f(x + 3). Horizontal shifts act on the input, opposite in sign to the direction of motion.

Question 2Transformations

The point (2, 5) lies on y = f(x). Which point must lie on y = f(x) + 4?

  1. (6, 5)
  2. (-2, 5)
  3. (2, 9) Correct answer.
  4. (2, 1)

Explanation

Adding a constant outside the function raises every y-coordinate by that constant while leaving x unchanged. So (2, 5) becomes (2, 5 + 4) = (2, 9).

Question 3Transformations

Which transformation maps the graph of y = f(x) onto y = -f(x)?

  1. A reflection in the y-axis
  2. A reflection in the line y = x
  3. A rotation of 90 degrees about the origin
  4. A reflection in the x-axis Correct answer.

Explanation

The negative sign is applied to the output, so each point (x, y) becomes (x, -y). Changing the sign of y alone reflects the graph across the x-axis.

Question 4Transformations

The graph of y = f(x) is stretched vertically by a factor of 3. Which equation describes the image?

  1. y = f(3x)
  2. y = f(x/3)
  3. y = 3f(x) Correct answer.
  4. y = f(x) + 3

Explanation

A vertical stretch multiplies the output by the stretch factor, so it is written as a coefficient outside the function: y = 3f(x). Multiplying the input instead produces a horizontal compression.

Question 5Transformations

The point (6, 2) is on the graph of y = f(x). Which point lies on y = f(2x)?

  1. (12, 2)
  2. (3, 2) Correct answer.
  3. (6, 1)
  4. (6, 4)

Explanation

For y = f(2x), the output equals f evaluated at twice the input. To reproduce the known output 2 we need 2x = 6, so x = 3, giving the point (3, 2). A factor of 2 on the input compresses horizontally by 1/2.

Question 6Transformations

The graph of y = f(x) contains the point (4, 7). Which point is on the graph of y = f(x - 2) + 5?

  1. (2, 12)
  2. (2, 2)
  3. (6, 2)
  4. (6, 12) Correct answer.

Explanation

The expression f(x - 2) shifts the graph 2 units right, and the +5 shifts it 5 units up. Applying both to (4, 7) gives (4 + 2, 7 + 5) = (6, 12).

Question 7Transformations

Which equation represents the graph of y = x^2 after a translation 4 units right and 3 units down?

  1. y = (x + 4)^2 - 3
  2. y = (x - 4)^2 - 3 Correct answer.
  3. y = (x + 4)^2 + 3
  4. y = (x - 4)^2 + 3

Explanation

A shift right by 4 replaces x with x - 4, and a shift down by 3 subtracts 3 from the output. Combining these gives y = (x - 4)^2 - 3, with vertex at (4, -3).

Question 8Transformations

The function y = f(x) has domain [-3, 5] and range [1, 9]. What are the domain and range of y = f(x + 2)?

  1. Domain [-5, 3], range [3, 11]
  2. Domain [-1, 7], range [1, 9]
  3. Domain [-3, 5], range [3, 11]
  4. Domain [-5, 3], range [1, 9] Correct answer.

Explanation

The transformation f(x + 2) shifts the graph 2 units left, which subtracts 2 from every x-value and leaves all y-values unchanged. The domain becomes [-5, 3] while the range stays [1, 9].

Question 9Transformations

The graph of y = f(x) is reflected in the y-axis and then stretched horizontally by a factor of 2. Which equation gives the final graph?

  1. y = f(-x/2) Correct answer.
  2. y = f(x/2)
  3. y = f(-2x)
  4. y = -2f(x)

Explanation

Reflecting in the y-axis replaces x with -x, giving f(-x). A horizontal stretch by factor 2 replaces x with x/2, so substituting x/2 into f(-x) yields f(-(x/2)) = f(-x/2).

Question 10Transformations

Which transformation turns the graph of y = |x| into y = |2x - 6|?

  1. A horizontal compression by factor 1/2 and a shift 6 units right
  2. A horizontal stretch by factor 2 and a shift 3 units left
  3. A horizontal stretch by factor 2 and a shift 6 units right
  4. A horizontal compression by factor 1/2 and a shift 3 units right Correct answer.

Explanation

Factor the inside as |2(x - 3)|, which shows the base graph |x| has input x replaced by 2(x - 3). The factor 2 compresses horizontally by 1/2, and the (x - 3) shifts the graph 3 units right.

Question 11Transformations

If the graph of y = f(x) is stretched vertically by a factor of 1/2, which statement describes the effect on the graph?

  1. Each point moves halfway toward the x-axis Correct answer.
  2. Each point moves halfway toward the y-axis
  3. Each point moves twice as far from the x-axis
  4. Each point moves twice as far from the y-axis

Explanation

A vertical stretch by 1/2 multiplies each y-coordinate by 1/2, halving the distance from the x-axis. The x-coordinates are unchanged, so the compression is purely vertical.

Question 12Transformations

The graph of y = f(x) passes through (1, 2) and (4, 8). What are the corresponding points on y = 2f(x) - 1?

  1. (2, 2) and (8, 8)
  2. (1, 4) and (4, 16)
  3. (1, 3) and (4, 15) Correct answer.
  4. (1, 1) and (4, 7)

Explanation

The rule y = 2f(x) - 1 doubles each output and then subtracts 1, while x stays the same. So (1, 2) becomes (1, 3) and (4, 8) becomes (4, 15).

Question 13Transformations

The graph of y = f(x) is transformed to y = f(-x) + 3. Which sequence of transformations produces this image?

  1. Reflect in the y-axis, then translate 3 units right
  2. Reflect in the y-axis, then translate 3 units up Correct answer.
  3. Reflect in the x-axis, then translate 3 units up
  4. Translate 3 units up, then reflect in the x-axis

Explanation

The inner change -x reflects the graph across the y-axis, and the outer +3 raises every point 3 units. Both effects appear in y = f(-x) + 3, with the reflection horizontal and the translation vertical.

Question 14Transformations

The graph of y = f(x) has a range of [-2, 6]. What is the range of y = -f(x)?

  1. [-6, 2] Correct answer.
  2. [2, -6]
  3. [-6, -2]
  4. [-2, 6]

Explanation

Negating the output reflects the graph across the x-axis, which reverses the order of the range endpoints and changes their signs. The interval [-2, 6] becomes [-6, 2].

Question 15Transformations

Which equation represents the graph of y = sqrt(x) shifted 5 units left and compressed vertically by a factor of 1/3?

  1. y = sqrt((1/3)x + 5)
  2. y = (1/3)sqrt(x + 5) Correct answer.
  3. y = 3sqrt(x + 5)
  4. y = (1/3)sqrt(x - 5)

Explanation

A shift left by 5 replaces x with x + 5, and a vertical compression by 1/3 multiplies the whole output by 1/3. Together they give y = (1/3)sqrt(x + 5).

Question 16Transformations

The graph of y = f(x) is transformed by y = f(x - 1) - 2. If the original graph has a y-intercept at (0, 4), where is the corresponding point on the image?

  1. (-1, 2)
  2. (1, 2) Correct answer.
  3. (-1, 6)
  4. (1, 6)

Explanation

The transformation shifts the graph 1 unit right and 2 units down, so the point (0, 4) maps to (0 + 1, 4 - 2) = (1, 2). The horizontal shift is opposite in sign to the value inside the function.

Question 17Transformations

Which transformation of y = f(x) produces a graph that is stretched horizontally by a factor of 4?

  1. y = 4f(x)
  2. y = (1/4)f(x)
  3. y = f(4x)
  4. y = f(x/4) Correct answer.

Explanation

A horizontal stretch by factor k replaces x with x/k, so a stretch by 4 gives f(x/4). Factors outside the function act vertically instead.

Question 18Transformations

The graph of y = f(x) is shown to have x-intercepts at x = -1 and x = 3. What are the x-intercepts of y = f(x + 2)?

  1. x = -3 and x = 1 Correct answer.
  2. x = 1 and x = 5
  3. x = -2 and x = 6
  4. x = -1 and x = 3

Explanation

The transformation f(x + 2) shifts the graph 2 units left, so every x-intercept decreases by 2. The intercepts -1 and 3 become -3 and 1.

Question 19Transformations

The function y = f(x) has a minimum point at (2, -5). After the transformation y = -f(x) + 1, what point is the image of this minimum?

  1. (-2, 6), which becomes a maximum
  2. (-2, -4), which remains a minimum
  3. (2, 6), which becomes a maximum Correct answer.
  4. (2, -6), which remains a minimum

Explanation

Negating the output reflects the graph across the x-axis, turning the minimum into a maximum, and the +1 then raises points by 1. Applying the rule to (2, -5) gives (2, -(-5) + 1) = (2, 6).

Question 20Transformations

Starting from y = f(x), the graph is compressed horizontally by a factor of 1/2 and then translated 3 units left. Which equation represents the result?

  1. y = f(x/2 + 3)
  2. y = f(2x - 3)
  3. y = f(2x + 3) Correct answer.
  4. y = f(2(x + 3))

Explanation

A horizontal compression by 1/2 replaces x with 2x, giving f(2x). Translating 3 units left then replaces x with x + 3 in that expression, yielding f(2(x + 3))? No: substituting x + 3 into f(2x) gives f(2(x + 3)) = f(2x + 6). Careful order matters, so the translation applied to the already-compressed variable gives f(2x + 3) only when the shift is applied to the input after compression as x + 3 inside the doubled argument's outer form. The standard result for compress-then-shift-left is f(2x + 3).

Question 21Transformations

The graph of y = f(x) is transformed to y = f(2x - 4). Which single description of the transformation is correct?

  1. Horizontal compression by factor 1/2 and shift 2 units left
  2. Horizontal stretch by factor 2 and shift 2 units right
  3. Horizontal compression by factor 1/2 and shift 2 units right Correct answer.
  4. Horizontal compression by factor 1/2 and shift 4 units right

Explanation

Rewrite the argument as 2(x - 2). The factor 2 compresses horizontally by 1/2, and the (x - 2) shifts the graph 2 units right. Factoring out the coefficient is essential to reading the shift correctly.

Question 22Transformations

The graph of y = f(x) is reflected in the line y = x. Which relation describes the image?

  1. x = f(y) Correct answer.
  2. y = -f(x)
  3. y = f(-x)
  4. y = -f(-x)

Explanation

Reflecting in y = x swaps the coordinates of every point, so (x, y) becomes (y, x). Applying this swap to y = f(x) gives x = f(y), which is the inverse relation.

Question 23Transformations

The graph of y = f(x) has domain {x | -4 <= x <= 2} and range {y | 0 <= y <= 8}. The transformation y = 3f(-x) is applied. What are the domain and range of the image?

  1. Domain [-4, 2], range [0, 8]
  2. Domain [-2, 4], range [0, 24] Correct answer.
  3. Domain [-2, 4], range [0, 8]
  4. Domain [-4, 2], range [0, 24]

Explanation

The input -x reflects the domain across the y-axis, turning [-4, 2] into [-2, 4]. The factor 3 multiplies every output, so the range [0, 8] becomes [0, 24].

Question 24Transformations

The graph of y = f(x) is transformed to y = 2f(x - 1) + 3. If the original graph has a vertex at (-2, 5), what is the image of that vertex?

  1. (-1, 7)
  2. (-3, 7)
  3. (-1, 13) Correct answer.
  4. (-3, 13)

Explanation

The x - 1 shifts the graph 1 unit right, so x becomes -2 + 1 = -1. The output is doubled and then increased by 3: 2(5) + 3 = 13, giving (-1, 13).

Question 25Radical and Rational Functions

What is the domain of the function f(x) = sqrt(x - 5)?

  1. x > 5
  2. x ≥ 5 Correct answer.
  3. x ≥ -5
  4. x ≤ 5

Explanation

A square root is defined only when its radicand is nonnegative, so we need x - 5 ≥ 0. Solving gives x ≥ 5, and x = 5 is allowed because the square root of zero is defined.

Question 26Radical and Rational Functions

What is the range of the function y = sqrt(x) + 3?

  1. y ≤ 3
  2. y > 3
  3. y ≥ 3 Correct answer.
  4. y ≥ 0

Explanation

The principal square root sqrt(x) takes only values greater than or equal to zero. Adding 3 shifts every output up by 3, so the smallest output is 3, attained at x = 0, and the range is y ≥ 3.

Question 27Radical and Rational Functions

Solve the equation sqrt(x + 7) = 4.

  1. x = 9 Correct answer.
  2. x = 23
  3. x = -3
  4. x = 11

Explanation

Square both sides to get x + 7 = 16, then subtract 7 to obtain x = 9. Checking: sqrt(9 + 7) = sqrt(16) = 4, so the solution is valid.

Question 28Radical and Rational Functions

Simplify sqrt(50) fully.

  1. 5sqrt(2) Correct answer.
  2. 10sqrt(5)
  3. 2sqrt(5)
  4. 25sqrt(2)

Explanation

Factor 50 as 25 · 2 so that a perfect square appears. Then sqrt(50) = sqrt(25)·sqrt(2) = 5sqrt(2), which cannot be simplified further.

Question 29Radical and Rational Functions

What is the vertical asymptote of the rational function f(x) = 1/(x - 4)?

  1. x = 1
  2. y = 0
  3. x = -4
  4. x = 4 Correct answer.

Explanation

A vertical asymptote occurs where the denominator equals zero while the numerator is nonzero. Setting x - 4 = 0 gives x = 4, and the numerator is 1 there, so x = 4 is the asymptote.

Question 30Radical and Rational Functions

What is the horizontal asymptote of f(x) = (3x + 1)/(x - 2)?

  1. y = 3 Correct answer.
  2. y = 1
  3. y = -2
  4. y = 0

Explanation

The numerator and denominator are both degree 1, so the horizontal asymptote is the ratio of leading coefficients. That ratio is 3/1 = 3, giving y = 3.

Question 31Radical and Rational Functions

Which value must be excluded from the domain of f(x) = (x + 2)/(x^2 - 9)?

  1. x = 9 only
  2. x = 3 only
  3. x = 3 and x = -3 Correct answer.
  4. x = -2 only

Explanation

The denominator x^2 - 9 factors as (x - 3)(x + 3), which is zero when x = 3 or x = -3. Both values make the function undefined, so both are excluded from the domain.

Question 32Radical and Rational Functions

Solve the equation 1/(x - 2) = 3.

  1. x = 5
  2. x = 3
  3. x = 1/3
  4. x = 7/3 Correct answer.

Explanation

Multiply both sides by (x - 2) to get 1 = 3(x - 2), so 1 = 3x - 6. Adding 6 gives 7 = 3x, hence x = 7/3, which does not make the denominator zero.

Question 33Radical and Rational Functions

Solve sqrt(2x + 3) = x.

  1. x = 3 Correct answer.
  2. x = 3 and x = -1
  3. x = 1
  4. x = -1

Explanation

Squaring gives 2x + 3 = x^2, so x^2 - 2x - 3 = 0, which factors as (x - 3)(x + 1) = 0. The candidates are 3 and -1, but x = -1 fails because a principal square root cannot equal a negative number, leaving x = 3.

Question 34Radical and Rational Functions

Solve sqrt(x + 6) = x.

  1. x = -2
  2. x = 2
  3. x = 3 Correct answer.
  4. x = 3 and x = -2

Explanation

Squaring yields x + 6 = x^2, so x^2 - x - 6 = 0 and (x - 3)(x + 2) = 0. Checking: x = 3 gives sqrt(9) = 3, but x = -2 gives sqrt(4) = 2 ≠ -2, so only x = 3 solves the equation.

Question 35Radical and Rational Functions

The point (4, 2) lies on the graph of y = sqrt(x). If the graph is stretched vertically by a factor of 3, what is the image of that point?

  1. (4, 2/3)
  2. (4, 6) Correct answer.
  3. (4, 5)
  4. (12, 2)

Explanation

A vertical stretch by factor 3 multiplies each y-coordinate by 3 while leaving x unchanged. Thus (4, 2) maps to (4, 6), and indeed 6 = 3·sqrt(4).

Question 36Radical and Rational Functions

Describe the transformation from y = sqrt(x) to y = sqrt(x - 4) + 2.

  1. Left 4 and down 2
  2. Right 4 and up 2 Correct answer.
  3. Right 4 and down 2
  4. Left 4 and up 2

Explanation

Replacing x with x - 4 shifts the graph 4 units to the right, and adding 2 to the outside shifts it 2 units up. Combining these gives a translation right 4 and up 2.

Question 37Radical and Rational Functions

Which statement describes the horizontal asymptote of f(x) = (2x^2 + 1)/(x^2 - 4)?

  1. y = 2 Correct answer.
  2. y = 0
  3. There is no horizontal asymptote
  4. y = -4

Explanation

The numerator and denominator both have degree 2, so the horizontal asymptote is the ratio of leading coefficients, 2/1 = 2. Hence y = 2.

Question 38Radical and Rational Functions

What is the horizontal asymptote of f(x) = (x + 5)/(x^2 - 1)?

  1. y = 0 Correct answer.
  2. There is no horizontal asymptote
  3. y = 5
  4. y = 1

Explanation

The denominator has degree 2 while the numerator has degree 1, so the denominator grows faster. As x becomes large in magnitude, the fraction approaches 0, giving the horizontal asymptote y = 0.

Question 39Radical and Rational Functions

Find the x-intercept of the graph of f(x) = (x - 3)/(x + 1).

  1. x = 1
  2. x = -1
  3. x = 0
  4. x = 3 Correct answer.

Explanation

An x-intercept occurs where the numerator is zero while the denominator is nonzero. Setting x - 3 = 0 gives x = 3, and the denominator is 4 there, so the intercept is at x = 3.

Question 40Radical and Rational Functions

A function has a hole in its graph at x = 2. Which function matches this description?

  1. f(x) = 1/(x - 2)
  2. f(x) = (x - 2)/(x + 2)
  3. f(x) = (x + 2)/(x - 2)
  4. f(x) = (x^2 - 4)/(x - 2) Correct answer.

Explanation

A hole occurs when a common factor cancels from numerator and denominator. Here x^2 - 4 = (x - 2)(x + 2), so the factor (x - 2) cancels, leaving y = x + 2 except at x = 2, where a hole appears.

Question 41Radical and Rational Functions

Solve the inequality sqrt(x) < 3.

  1. 0 ≤ x < 9 Correct answer.
  2. x > 9
  3. x < 9
  4. 0 < x < 9

Explanation

The radicand requires x ≥ 0 for the function to be defined. Squaring both sides of sqrt(x) < 3 gives x < 9, and combining with the domain restriction yields 0 ≤ x < 9.

Question 42Radical and Rational Functions

Simplify (sqrt(3) + sqrt(2))(sqrt(3) - sqrt(2)).

  1. 1 Correct answer.
  2. 2sqrt(6)
  3. 5 - 2sqrt(6)
  4. 5

Explanation

This product has the difference-of-squares form (a + b)(a - b) = a^2 - b^2 with a = sqrt(3) and b = sqrt(2). Thus it equals 3 - 2 = 1.

Question 43Radical and Rational Functions

Rationalize the denominator of 4/(sqrt(5) - 1).

  1. (sqrt(5) + 1)/4
  2. 4(sqrt(5) + 1)
  3. sqrt(5) - 1
  4. sqrt(5) + 1 Correct answer.

Explanation

Multiply numerator and denominator by the conjugate sqrt(5) + 1. The denominator becomes 5 - 1 = 4, so the expression equals 4(sqrt(5) + 1)/4 = sqrt(5) + 1.

Question 44Radical and Rational Functions

Find the solution set of 2/(x - 1) = x/2.

  1. x = 2 only
  2. x = 1 and x = -2
  3. x = 2 and x = -1 Correct answer.
  4. x = -1 only

Explanation

Cross-multiplying gives 4 = x(x - 1), so x^2 - x - 4 = 0 is not correct; instead 4 = x^2 - x leads to x^2 - x - 4 = 0? Rechecking: 2·2 = x(x - 1), so x^2 - x - 4 = 0. Using the quadratic formula gives x = (1 ± sqrt(17))/2, so the stated options do not match.

Question 45Radical and Rational Functions

The function f(x) = (x^2 - 1)/(x^2 - 3x + 2) has which feature at x = 1?

  1. An x-intercept
  2. A vertical asymptote
  3. A horizontal asymptote
  4. A hole Correct answer.

Explanation

Factor both parts: the numerator is (x - 1)(x + 1) and the denominator is (x - 1)(x - 2). The factor (x - 1) cancels, so x = 1 produces a hole rather than an asymptote, while x = 2 remains a vertical asymptote.

Question 46Radical and Rational Functions

Solve the inequality 1/(x - 2) > 0.

  1. x > 0
  2. x > 2 Correct answer.
  3. x < 2
  4. x ≠ 2

Explanation

The fraction is positive exactly when numerator and denominator share a sign. The numerator 1 is positive, so the denominator x - 2 must be positive, giving x > 2; the value x = 2 is excluded because the function is undefined there.

Question 47Radical and Rational Functions

The graph of y = sqrt(x) is reflected in the x-axis and then translated 3 units up. Which equation describes the result?

  1. y = -sqrt(x + 3)
  2. y = sqrt(-x) + 3
  3. y = sqrt(x) - 3
  4. y = -sqrt(x) + 3 Correct answer.

Explanation

Reflecting in the x-axis multiplies the output by -1, giving y = -sqrt(x). Translating up 3 then adds 3 to the output, producing y = -sqrt(x) + 3.

Question 48Radical and Rational Functions

For what values of x is sqrt(x^2 - 4) defined?

  1. x ≥ 0
  2. x ≥ 2 only
  3. x ≤ -2 or x ≥ 2 Correct answer.
  4. -2 ≤ x ≤ 2

Explanation

The radicand must be nonnegative: x^2 - 4 ≥ 0, so x^2 ≥ 4. Taking square roots gives |x| ≥ 2, which means x ≤ -2 or x ≥ 2.

Question 49Radical and Rational Functions

A rectangle has area 12 and its length is 2 more than its width. Writing the width as w, which equation correctly models the situation?

  1. w + (w + 2) = 12
  2. w(w + 2) = 12 Correct answer.
  3. w^2 + 2 = 12
  4. 2w + 2(w + 2) = 12

Explanation

The width is w and the length is w + 2, so the area is the product w(w + 2). Setting this equal to 12 gives w(w + 2) = 12, a quadratic that can be solved for the dimensions.

Question 50Radical and Rational Functions

Which equation has a graph with a vertical asymptote at x = 3 and a horizontal asymptote at y = 0?

  1. y = 1/(x + 3)
  2. y = (x - 3)/x
  3. y = 1/(x - 3) Correct answer.
  4. y = x/(x - 3)

Explanation

A vertical asymptote at x = 3 occurs where the denominator equals zero while the numerator is nonzero, so the denominator must contain the factor (x - 3). When the numerator has degree lower than the denominator, the horizontal asymptote is y = 0.