Key Features of Graphs Worksheet Answer Key – Algebra 1

Problem 1 Intercepts of a line

y-intercept: x = 0 gives y = −6. x-intercept: 0 = 2x − 6 gives x = 3.

(3, 0) and (0, −6)

Problem 2 Intercepts with a negative slope

x = 0: y = 12. y = 0: 3x = 12, so x = 4.

(4, 0) and (0, 12)

Problem 3 Increasing or decreasing

The slope is −4, which is negative, so outputs fall as x increases.

Decreasing

Problem 4 Rate from a table

Outputs grow, so it is increasing. Rate = (28 − 16) ÷ (6 − 2) = 12 ÷ 4 = 3.

Increasing; average rate of change 3

Problem 5 Reading a parabola

The graph crosses the x-axis at −1 and 3, crosses the y-axis at −3 and its lowest point is (1, −4).

x-intercepts −1 and 3; y-intercept −3; minimum (1, −4)

Problem 6 Positive and negative intervals

The graph is above the x-axis to the left of −1 and to the right of 3, and below it between −1 and 3.

Positive: x < −1 or x > 3. Negative: −1 < x < 3.

Problem 7 Increasing, decreasing, extremes

The graph rises from x = −5 to −3, falls from −3 to 1, and rises from 1 to 5.

Increasing: −5 < x < −3 and 1 < x < 5. Decreasing: −3 < x < 1. Relative max (−3, 3); relative min (1, −1).

Problem 8 Rate of change of a quadratic

f(2) = 4 − 12 + 5 = −3. f(6) = 36 − 36 + 5 = 5.

Rate = (5 − (−3)) ÷ (6 − 2) = 8 ÷ 4 = 2.

2

Problem 9 Rate with a negative input

f(−4) = 16 − 11 = 5. f(3) = 9 − 11 = −2.

Rate = (−2 − 5) ÷ (3 − (−4)) = −7 ÷ 7 = −1.

−1

Problem 10 Rate from a graph

Points (2, 2) and (6, 18). Slope = (18 − 2) ÷ (6 − 2) = 16 ÷ 4 = 4.

4

Problem 11 End behavior

(a) A line with negative slope: as x → ∞, f(x) → −∞, and as x → −∞, f(x) → +∞.

(b) A parabola opening down: g(x) → −∞ at both ends.

(a) falls to the right, rises to the left (b) falls at both ends

Problem 12 Symmetry of a parabola

The point 0 is 3 units left of the axis, so the mirror point is 3 units right: x = 6.

Check: f(0) = 9 − 2 = 7 and f(6) = 9 − 2 = 7.

(6, 7)

Problem 13 Compare an equation and a table

f has rate 4. g increases by 5 each step, and g(0) = 1, so g(x) = 5x + 1.

g has the greater rate (5 versus 4); g(x) = 5x + 1

Problem 14 Balloon flight

Increasing 0 to 10 min, constant 10 to 20 min, decreasing 20 to 35 min. The greatest altitude is 600 ft.

From 0 to 10: (600 − 0) ÷ 10 = 60 ft/min. From 20 to 35: (0 − 600) ÷ 15 = −40 ft/min.

Max 600 ft; +60 ft/min and −40 ft/min

Problem 15 Bike trip table

0 to 2: 30 ÷ 2 = 15 mph. 2 to 4: (60 − 30) ÷ 2 = 15 mph. 1 to 3: (44 − 14) ÷ 2 = 15 mph.

0 to 1: 14 ÷ 1 = 14 mph, which differs from 15.

15, 15, 15 and 14 mph: not constant

Problem 16 Stock price

0 to 4: (60 − 48) ÷ 4 = 3 dollars per week (rising).

4 to 10: (42 − 60) ÷ 6 = −3 dollars per week (falling).

0 to 10: (42 − 48) ÷ 10 = −0.6 dollars per week overall.

+3, −3 and −0.6 dollars per week

Problem 17 Two phone plans

Plan B increases by 8 per 100 minutes, so C = 20 + 0.08m (check: 20 + 8 = 28). A’s rate 0.12 is greater than B’s 0.08.

At 500 minutes: A = 18 + 60 = 78 and B = 20 + 40 = 60.

Equal when 18 + 0.12m = 20 + 0.08m, so 0.04m = 2 and m = 50.

Plan A grows faster; $78 versus $60; equal at 50 minutes

Problem 18 Fish tank

Decreasing from 0 to 6 min, constant (10 in) from 6 to 10 min, increasing from 10 to 16 min. Lowest level: 10 inches.

0 to 6: (10 − 40) ÷ 6 = −5 in/min. 10 to 16: (40 − 10) ÷ 6 = 5 in/min.

Min 10 in; −5 in/min and +5 in/min

Problem 19 Cupcake profit

Zeros: x = 2 and x = 10 (break-even points). Maximum: P(6) = −72 + 144 − 40 = 32, so $3,200 at 600 cupcakes. y-intercept: P(0) = −40, a loss of $4,000 before any sale. Positive for 2 < x < 10.

Average rate from 2 to 6: (32 − 0) ÷ 4 = 8.

Zeros 2 and 10; max 32 at x = 6; y-intercept −40; positive for 2 < x < 10

Problem 20 Compare two functions

f has its maximum at (3, 9) and is positive for 0 < x < 6, a width of 6.

g: vertex at x = 2, g(2) = −4 + 8 + 8 = 12, so maximum 12. Its zeros are x = 2 ± √12 ≈ −1.46 and 5.46, a width ≈ 6.93.

g has the larger maximum (12 versus 9) and the wider positive interval (about 6.93 versus 6)