precalc apmc

[Poly 1] f is a V-shaped graph with its vertex at the origin, f(x) = abs(x).
What is the average rate of change of f on −3 ≤ x ≤ 3?
(A) −1
(B) 0
(C) 1
(D) Undefined

(B) 0
f(−3) = 3 and f(3) = 3, so (3 − 3)/(3 − (−3)) = 0/6 = 0.

[Poly 2] Which function has lim x→−∞ p(x) = ∞ and lim x→∞ p(x) = ∞?
(A) (x − 10)(x + 2)(x + 3)
(B) (x − 10)(x + 2)(x + 3)(x + 4)
(C) (10 − x)(x + 2)(x + 3)
(D) (10 − x)(x + 2)(x + 3)(x + 4)

(B)
For both ends to go to +∞, the degree must be even and the leading coefficient positive.
(B) has degree 4 and leading coefficient +1.
(A) and (C) have odd degree, and (D) has leading coefficient −1.

[Poly 3] f(x) = ax^b, where a is an integer and b is a positive integer.
lim x→−∞ f(x) = −∞ and lim x→∞ f(x) = −∞. Which must be true?
(A) a must be positive, because the end behaviors are the same
(B) a must be negative, because the end behaviors are different
(C) b must be even, because the end behaviors are the same
(D) b must be odd, because the end behaviors are different

(C)
Both ends go to −∞, so the end behaviors are the same, and that means b is even.
(B) gets the sign of a right but gives the wrong reason.

[Poly 4] For polynomial f, lim x→−∞ f(x) = ∞ and lim x→∞ f(x) = −∞.
Which must be true?
(A) Even degree, negative leading coefficient
(B) Even degree, positive leading coefficient
(C) Odd degree, negative leading coefficient
(D) Odd degree, positive leading coefficient

(C)
The ends go in opposite directions, so the degree is odd.
The graph is high on the left and low on the right, so the leading coefficient is negative.

[Poly 5] For polynomial g, lim x→−∞ g(x) = −∞.
Which could define g(x)?
(A) −5x − 2x^4
(B) −5x^2 − 2x^7
(C) −9000x − x^5/5
(D) −x^3 + 2x^4

(A)
The leading term −2x^4 has even degree and a negative coefficient, so g → −∞ as x → −∞.
The leading terms of (B) and (C) are odd degree with negative coefficients, so they go to +∞ on the left.
(D) has +2x^4, which also goes to +∞.

[Poly 6] f(x) = −2x^7 + 5x^4 + 6x^2 − 3.
Which describes the end behavior of f as the input values increase without bound?
(A) lim x→∞ f(x) = ∞
(B) lim x→∞ f(x) = −∞
(C) lim x→−∞ f(x) = ∞
(D) lim x→−∞ f(x) = −∞

(B)
The leading term −2x^7 controls the end behavior, and it goes to −∞ as x → ∞.
(C) is also a true statement, but it describes x → −∞, not increasing inputs.

[Poly 7] (Calculator) R(d) = (1/200)(−d^4 + 35d^3 − 411d^2 + 1845d − 2686.5) is the rate of change of M, the number of museum visitors on day d.
At which values of d does y = M(d) have a point of inflection?
(A) 8.627 only
(B) 3.894 and 13.728 only
(C) 5.911 and 11.590
(D) 3.894, 8.627, and 13.728

(D)
R is the rate of change of M, so M has an inflection point wherever R switches between increasing and decreasing.
Those are R's local extrema.
Graphing R shows extrema at d ≈ 3.894, 8.627, and 13.728.

[Poly 8] (Graph in packet: polynomial g has a tall peak at about y = 3.5 between x = 0 and x = 2, a small dip below the axis between x = 2 and x = 3, and a small hump just before x = 3, then it drops off.)
f is defined for 0 ≤ x ≤ 3 and is identical to g there.
How many total local minima and local maxima does f have?
(A) Two
(B) Four
(C) Five
(D) Seven

(C) Five
Inside the interval there are 3: the tall peak (max), the dip (min), and the small hump (max).
On a restricted domain the endpoints also count.
f rises right after x = 0 and falls into x = 3, so both endpoints are local minima.
3 + 2 = 5.

[Poly 9] f(x) = ax^4 + bx^3 + cx^2 + dx + k, where a ≠ 0.
Which is true?
(A) f has both a global max and a global min.
(B) f has either a global max or a global min, but not both.
(C) f has neither.
(D) It can't be determined without more info about b, c, d, and k.

(B)
A degree-4 polynomial has both ends going the same way.
If a > 0, both ends go to +∞, so f has a global min and no global max.
If a < 0, it has a global max and no global min.

[Poly 10] (Graph in packet: f rises from the bottom left to a high peak, dips to a local min on the y-axis, rises to a smaller peak, drops to a deep min below the x-axis, then rises steeply.)
How many points of inflection does the graph of f have on the shown portion?
(A) One
(B) Three
(C) Four
(D) Five

(B) Three
The graph is concave down at the first peak, up at the first min, down at the second peak, and up at the deep min.
That is 3 changes in concavity.

[Poly 11] f is linear.
x: −3, 3, 6
f(x): −10, −2, 2
What is f(13)?
(A) 4
(B) 29/4
(C) 28/3
(D) 34/3

(D) 34/3
Slope = (−2 − (−10))/(3 − (−3)) = 8/6 = 4/3
f(13) = 2 + (4/3)(13 − 6) = 2 + 28/3 = 34/3

[Poly 12] (Calculator) A linear function P models used-car price in dollars by age t in years.
P(4) = 7300 and P(7) = 5500. Which is true?
(A) The price drops about 25% per year.
(B) The price drops about 75% per year.
(C) The price drops about $600 per year.
(D) The price drops about $1800 per year.

(C)
Slope = (5500 − 7300)/(7 − 4) = −1800/3 = −600
A linear model drops by a constant dollar amount each year, not a constant percent.

[Poly 13] For a constant a, the average rate of change of f from x = a to x = a + 1 is 2a + 1.
Which is true?
(A) The ARoC is positive, so f could be a line with positive slope.
(B) The ARoC is positive, so f could be a parabola that opens up.
(C) The ARoC increases at a constant rate, so f could be a line with positive slope.
(D) The ARoC increases at a constant rate, so f could be a parabola that opens up.

(D)
Each time a goes up by 1, 2a + 1 goes up by 2.
Rates of change that increase at a constant rate mean f is quadratic and opens up, for example f(x) = x^2.
A line would have constant rates of change.

[Poly 14] x: 5, 6, 7, 8, 9, 10
f(x): −17, −27, −39, −53, −69, −87
Which conclusion is consistent with the table?
(A) Concave up, because the ARoC is increasing
(B) Concave up, because the ARoC is decreasing
(C) Concave down, because the ARoC is increasing
(D) Concave down, because the ARoC is decreasing

(D)
The changes are −10, −12, −14, −16, −18.
The rates of change are decreasing, so f is concave down.

[Poly 15] Quadratic p has average rate of change −4 on 0 ≤ x ≤ 2 and −1 on 2 ≤ x ≤ 4.
What is the average rate of change of p on 6 ≤ x ≤ 8?
(A) 2
(B) 3
(C) 5
(D) It cannot be determined.

(C) 5
For a quadratic, rates of change over consecutive equal-length intervals change by a constant amount.
Going from −4 to −1 is +3, so:
[4, 6] → 2
[6, 8] → 5

[Poly 16] Time (s): 1, 3, 6, 11
Distance (m): 1, 9, 21, 41
If the pattern continues, what is the rate of change of the rates of change?
(A) 0 m/s, and the object is neither speeding up nor slowing down
(B) 0 m/s per second, and the object is neither speeding up nor slowing down
(C) 4 m/s, and the object is neither speeding up nor slowing down
(D) 4 m/s per second, and the object is speeding up

(B)
Rates: (9 − 1)/2 = 4, (21 − 9)/3 = 4, (41 − 21)/5 = 4 m/s
The speed is constant, so its rate of change is 0.
A rate of a rate has units of m/s per second.

[Poly 17] x: −2, −1, 0, 1
f(x): 5, 2, 1, 2
Which conclusion is consistent with the table?
(A) f could be linear; the rates of change are described by y = 2x.
(B) f could be linear; the rates of change are described by y = 2x + 1.
(C) f could be quadratic; the rates of change are described by y = 2x.
(D) f could be quadratic; the rates of change are described by y = 2x + 1.

(D)
The changes are −3, −1, 1. They go up by 2 each time, so f could be quadratic.
Plugging the left x of each interval into y = 2x + 1:
−2 → −3
−1 → −1
0 → 1

[Poly 18] g(x) = f(x + 1) − f(x)
h(x) = g(x + 1) − g(x)
If h(x) = −6 for all x, which must be true?
(A) The graphs of g and f always have negative slope.
(B) The graphs of g and f are concave down.
(C) g is decreasing, and the graph of f always has negative slope.
(D) g is decreasing, and the graph of f is concave down.

(D)
h gives the changes in g, and h is always negative, so g is decreasing.
g gives the rates of change of f, so f's rates of change are decreasing, which means f is concave down.
Nothing tells us f's slope is always negative.

[Unit 1.1-1.2 #1] Rate of change of f:
0 < x < 1: increasing
1 < x < 2: positive and constant
2 < x < 3: decreasing
3 < x < 4: negative and constant
On which interval is the graph of f concave down?
(A) 0 < x < 1
(B) 1 < x < 2
(C) 2 < x < 3
(D) 3 < x < 4

(C) 2 < x < 3
A graph is concave down where its rate of change is decreasing.

[Unit 1.1-1.2 #2] A toy car drives around a circular track that touches a wall at the start point.
Which graph shows its distance from the wall over time for 3 laps?
(A) Three smooth hills that start at 0 and return to 0
(B) One smooth wave that goes above and below the time axis
(C) A curve that keeps climbing in bumps
(D) Three straight-sided triangle peaks that start at 0 and return to 0

(A)
The distance starts at 0, smoothly grows to the diameter on the far side, then smoothly returns to 0, and this repeats 3 times.
(B) is out because distance can't be negative.
(C) is out because the distance doesn't keep growing.
(D) is out because the distance doesn't change at a constant rate around a circle.

[Unit 1.1-1.2 #3] (Graph in packet: f has a min at the origin, passes through (2, 16/3), peaks near x = 4 at about y = 10.7, then drops below the x-axis near x = 6.)
Which is true about the rate of change of f on 2 < x < 6?
(A) It is positive.
(B) It is negative.
(C) It is increasing.
(D) It is decreasing.

(D)
The graph is concave down on this whole interval. It starts steep upward, flattens at the peak, then falls more and more steeply.
f both increases and decreases here, so (A) and (B) are out.

[Unit 1.1-1.2 #4] (Calculator) A ball is thrown from a window. Its height is h(t) = −4.9t^2 + 4.4t + 15.24 meters.
Which describes the height?
(A) It starts at its max of 15.240 m and hits the ground 1.820 s after leaving the window.
(B) It starts at its max of 15.240 m and hits the ground 2.269 s after leaving the window.
(C) It rises to a max of 16.228 m and hits the ground 1.820 s after the max.
(D) It rises to a max of 16.228 m and hits the ground 2.269 s after the max.

(C)
Vertex: t = 4.4/9.8 ≈ 0.449 s, and h(0.449) ≈ 16.228 m.
h(t) = 0 at t ≈ 2.269 s.
Time from max to ground: 2.269 − 0.449 ≈ 1.820 s.
(D) is wrong because 2.269 s is measured from the throw, not from the max.

[Unit 1.1-1.2 #5] Rate of change of f:
1 < x < 2: positive and increasing
2 < x < 3: negative and increasing
3 < x < 4: positive and decreasing
4 < x < 5: negative and decreasing
What is true about f on 3 < x < 4?
(A) f is increasing and concave down.
(B) f is increasing and concave up.
(C) f is decreasing and concave down.
(D) f is decreasing and concave up.

(A)
A positive rate of change means f is increasing.
A decreasing rate of change means the graph is concave down.

[Unit 1.1-1.2 #6] Water is poured into a vase at a constant rate. The depth-vs-time graph is always increasing, concave up at first, then has a steady, steep increase.
Which vase fits?
(A) Hourglass: narrows to the middle, then widens
(B) Wide flat base with straight sides tapering in to a narrow straight neck
(C) Round bulb bottom with a narrow neck
(D) Narrow base that bulges out, then curves in to a narrow neck

(B)
Concave up means the depth rises faster and faster, so the vase must keep getting narrower going up, like B's tapered lower section.
The steady steep part is the narrow straight neck.
(C) and (D) widen first, which would make the graph start out concave down.

[Unit 1.1-1.2 #7] Height changes (ft) over 6-second intervals from t = 0 to 30:
Drone A: +17, −4, +11, −5, −3
Drone B: +5, +3, +3, +2, +3
Compare their average rates of change from t = 0 to t = 30.
(A) They are equal.
(B) Drone A's is greater.
(C) Drone B's is greater.
(D) They can't be determined without the actual heights.

(A)
Drone A: 17 − 4 + 11 − 5 − 3 = 16 ft
Drone B: 5 + 3 + 3 + 2 + 3 = 16 ft
Both average 16/30 ft per second. Only the total change matters, not the actual heights.

[Unit 1.1-1.2 #8] Average rates of change of f:
0 ≤ x ≤ 1: 10
1 ≤ x ≤ 4: −5
4 ≤ x ≤ 8: 2
8 ≤ x ≤ 10: 6
On which interval does f increase the most?
(A) 0 ≤ x ≤ 1
(B) 1 ≤ x ≤ 4
(C) 4 ≤ x ≤ 8
(D) 8 ≤ x ≤ 10

(D)
Total change = rate × interval length:
10 · 1 = 10
−5 · 3 = −15
2 · 4 = 8
6 · 2 = 12
The biggest increase is 12, on 8 ≤ x ≤ 10.

[Unit 1.1-1.2 #9] (Graph in packet: g touches 0 at x = −3, peaks near (−2, 4.6), touches 0 at x = 0, has a small hump near (1, 1), dips near x = 2.3, has another small hump near x = 3.4, then drops steeply below −4 just past x = 4.)
On which interval is the average rate of change of g least?
(A) −3 ≤ x ≤ −2
(B) −1 ≤ x ≤ 0
(C) 1 ≤ x ≤ 2
(D) 3 ≤ x ≤ 4

(D)
g falls from slightly above 0 at x = 3 to about −3 at x = 4. That is the steepest drop of the four intervals.
(A) is positive, and (B) and (C) fall less.

[Unit 1.1-1.2 #10] (Calculator) f(x) = x^2 + 3x − 5.
Which describes f?
(A) It always has a positive rate of change.
(B) It always has a negative rate of change.
(C) Positive rate of change for x < −1.5, negative for x > −1.5
(D) Negative rate of change for x < −1.5, positive for x > −1.5

(D)
The vertex is at x = −3/(2 · 1) = −1.5, and the parabola opens up.
So f decreases to the left of −1.5 and increases to the right.

[Unit 1.1-1.2 #11] (Graph in packet: river temperature is lowest, about 13°C, in early February, rises to a high of about 23°C on August 1, then falls.)
On the first day of which month is the rate of change of temperature greatest?
(A) February
(B) May
(C) August
(D) November

(B) May
The graph is climbing steeply around May.
February and August are near the min and max, where the rate is about 0.
In November the temperature is falling, so the rate is negative.

[Unit 1.1-1.2 #12] f has a negative average rate of change on every interval in 0 ≤ x ≤ 10.
g has a negative average rate of change on 0 ≤ x < 5 and a positive one on 5 < x ≤ 10.
What must be true about h(x) = f(x) + g(x)?
(A) h is decreasing on 0 ≤ x ≤ 10.
(B) h is decreasing on 0 ≤ x < 5 and increasing on 5 < x ≤ 10.
(C) h is decreasing on 0 ≤ x < 5 and neither increasing nor decreasing on 5 < x ≤ 10.
(D) h is decreasing on 0 ≤