12 multiple-choice questions, progressively harder.
The point (2,5)(2, 5)(2,5) lies on y=f(x)y = f(x)y=f(x). Which point must lie on y=f(x−4)+1y = f(x - 4) + 1y=f(x−4)+1?
Solution
Correct answer: B
The inside x−4x - 4x−4 shifts right 444 and the outside +1+1+1 shifts up 111.
(2+4, 5+1)=(6,6)(2 + 4,\ 5 + 1) = (6, 6)(2+4, 5+1)=(6,6)
The input gains 444 and the height gains 111.
The point (4,6)(4, 6)(4,6) lies on y=f(x)y = f(x)y=f(x). Which point must lie on y=f(2x)y = f(2x)y=f(2x)?
Correct answer: C
The point reappears where the inside 2x2x2x equals the old input 444.
2x=4 ⇒ x=2,height 62x = 4 \ \Rightarrow\ x = 2, \quad \text{height } 62x=4 ⇒ x=2,height 6
The input is halved and the height stays.
The point (−3,2)(-3, 2)(−3,2) lies on y=f(x)y = f(x)y=f(x). Which point must lie on y=−f(x)+5y = -f(x) + 5y=−f(x)+5?
Correct answer: D
The outside negates the height, then adds 555; the input is unchanged.
(−3, −(2)+5)=(−3,3)(-3,\ -(2) + 5) = (-3, 3)(−3, −(2)+5)=(−3,3)
Negate first to get −2-2−2, then −2+5=3-2 + 5 = 3−2+5=3.
The function fff has domain 2≤x≤82 \le x \le 82≤x≤8. What is the domain of y=f(x−3)y = f(x - 3)y=f(x−3)?
A shift right by 333 adds 333 to every input, moving both ends of the domain.
2+3≤x≤8+3 ⇒ 5≤x≤112 + 3 \le x \le 8 + 3 \ \Rightarrow\ 5 \le x \le 112+3≤x≤8+3 ⇒ 5≤x≤11
The domain slides right with the graph.
The function fff has range −2≤y≤6-2 \le y \le 6−2≤y≤6. What is the range of y=f(x)+4y = f(x) + 4y=f(x)+4?
Adding 444 outside raises every output by 444, moving both ends of the range.
−2+4≤y≤6+4 ⇒ 2≤y≤10-2 + 4 \le y \le 6 + 4 \ \Rightarrow\ 2 \le y \le 10−2+4≤y≤6+4 ⇒ 2≤y≤10
The range slides up with the graph.
Write y=f(x−2)+3y = f(x - 2) + 3y=f(x−2)+3 as a description of moves on y=f(x)y = f(x)y=f(x).
Correct answer: A
The inside x−2x - 2x−2 is a horizontal shift and the outside +3+3+3 is a vertical shift.
x−2⇒right 2,+3⇒up 3x - 2 \Rightarrow \text{right } 2, \qquad +3 \Rightarrow \text{up } 3x−2⇒right 2,+3⇒up 3
The minus inside moves right, and the plus outside moves up.
Which describes the graph of y=f(3x)y = f(3x)y=f(3x) compared with y=f(x)y = f(x)y=f(x)?
An inside factor 333 makes the inside reach old values at a third of the input.
3x=u ⇒ x=u33x = u \ \Rightarrow\ x = \tfrac{u}{3}3x=u ⇒ x=3u
Every input is divided by 333, pulling the graph toward the yyy-axis, a horizontal compression.
A function fff has domain −4≤x≤4-4 \le x \le 4−4≤x≤4. What is the domain of y=f(−x)y = f(-x)y=f(−x)?
The point at old input uuu reappears where the inside −x-x−x equals uuu, at x=−ux = -ux=−u.
−4≤−x≤4 ⇒ −4≤x≤4-4 \le -x \le 4 \ \Rightarrow\ -4 \le x \le 4−4≤−x≤4 ⇒ −4≤x≤4
Negating flips the interval end for end, but this interval is symmetric about 000, so it returns to itself.
The point (2,3)(2, 3)(2,3) lies on y=f(x)y = f(x)y=f(x). Which point must lie on y=f(−x)y = f(-x)y=f(−x)?
Negating the input reflects across the yyy-axis, flipping the input's sign while the height stays.
(−2, 3)(-2,\ 3)(−2, 3)
The height 333 is unchanged.
Which rule describes reflecting y=f(x)y = f(x)y=f(x) across the yyy-axis and then shifting up 333?
Reflecting across the yyy-axis negates the input, then +3+3+3 raises the height.
f(x) → f(−x) → f(−x)+3f(x) \ \to\ f(-x) \ \to\ f(-x) + 3f(x) → f(−x) → f(−x)+3
The outside +3+3+3 shifts up; −f(x)-f(x)−f(x) would be the wrong reflection.
Compared with y=f(x)y = f(x)y=f(x), the graph of y=f(x)−6y = f(x) - 6y=f(x)−6 moves the range how?
Subtracting 666 outside lowers every output by 666, so the whole range drops.
y=f(x)−6 ⇒ range down 6y = f(x) - 6 \ \Rightarrow\ \text{range down } 6y=f(x)−6 ⇒ range down 6
Both endpoints of the range decrease by 666.
The point (9,2)(9, 2)(9,2) lies on y=f(x)y = f(x)y=f(x). Which point must lie on y=f(3x)y = f(3x)y=f(3x)?
The point reappears where the inside 3x3x3x equals the old input 999.
3x=9 ⇒ x=3,height 23x = 9 \ \Rightarrow\ x = 3, \quad \text{height } 23x=9 ⇒ x=3,height 2
The input is divided by 333 and the height stays.
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