Graphs and Solutions Worksheet Answer Key – Algebra 1

Problem 1 Test Two Points on a Line

(3, 11): 2(3) + 5 = 11, so it is on the graph. (−1, 4): 2(−1) + 5 = 3, not 4, so it is not.

(3, 11) is on the graph; (−1, 4) is not

Problem 2 Test Three Points on a Parabola

(4, 10): 16 − 6 = 10 ✓. (−3, 3): 9 − 6 = 3 ✓. (2, −1): 4 − 6 = −2, not −1.

(4, 10) and (−3, 3)

Problem 3 Complete the Solution Pair

1. 3(4) − 7 = 5.
2. (−3)² + 2(−3) = 9 − 6 = 3.

(a) (4, 5); (b) (−3, 3)

Problem 4 Growth y-Intercept and a Value

At x = 0: 5(1) = 5, so the intercept is (0, 5). At x = 3: 5(8) = 40. Since 2 > 1 the graph rises.

(0, 5); 40; rises

Problem 5 Decay y-Intercept and a Value

At x = 0: 80. At x = 3: 80 · 18 = 10. The values halve each step and never reach 0, so the asymptote is y = 0.

(0, 80); 10; y = 0

Problem 6 Negative Inputs

3⁻¹ = 13, so y = 6 · 13 = 2. 3⁻² = 19, so y = 6 · 19 = 23.

The values shrink toward 0 as x decreases.

2 and 23

Problem 7 Find Intersections from a Graph

Set equal: x² − 4 = x − 2, so x² − x − 2 = 0 and (x − 2)(x + 1) = 0. Then x = 2 or x = −1.

y = x − 2 gives 0 and −3.

(2, 0) and (−1, −3)

Problem 8 Substitution with y = x²

x² = 2x + 8, so x² − 2x − 8 = 0 and (x − 4)(x + 2) = 0.

x = 4 gives y = 16. x = −2 gives y = 4.

(4, 16) and (−2, 4)

Problem 9 System with a Missing Constant Term

x² − x − 6 = 2x − 6, so x² − 3x = 0 and x(x − 3) = 0.

x = 0 gives y = −6. x = 3 gives y = 0.

(0, −6) and (3, 0)

Problem 10 Exactly One Solution

x² − 6x + 10 = 2x − 6, so x² − 8x + 16 = 0 and (x − 4)² = 0. Then x = 4 and y = 2.

The line touches the parabola at one point only, so it is tangent to it.

(4, 2); the line is tangent to the parabola

Problem 11 No Solution

x² + 3 = 2x, so x² − 2x + 3 = 0. The discriminant is (−2)² − 4(1)(3) = 4 − 12 = −8, which is negative.

Therefore the graphs never meet.

No real solution (discriminant −8)

Problem 12 Tangent Condition

Set x² = 2x + c, so x² − 2x − c = 0. One solution needs discriminant 0: 4 + 4c = 0, so c = −1.

Then x² − 2x + 1 = 0, so x = 1 and y = 1.

c = −1; the point (1, 1)

Problem 13 Read an Exponential Graph

The curve crosses the y-axis at (0, 4). The values halve as x increases: 4, 2, 1, so b = 12.

As x → ∞ the values approach 0; as x → −∞ they grow without bound. The asymptote is y = 0.

(0, 4); y = 0; → 0 and → ∞; y = 4(12)^x

Problem 14 Rewrite with the Power Rule

1.08^t = (1.08^1/12)^12t. The twelfth root of 1.08 is about 1.0064.

So 100(1.08)^t ≈ 100(1.0064)^12t, a growth of about 0.64% per month.

Monthly factor ≈ 1.0064

Problem 15 Ball Over a Ramp

−x² + 6x = x + 4, so x² − 5x + 4 = 0 and (x − 1)(x − 4) = 0. Then x = 1 or x = 4.

Heights: h = 1 + 4 = 5 and h = 4 + 4 = 8.

(1 ft, 5 ft) and (4 ft, 8 ft)

Problem 16 Solve and Check Both Equations

x² − 2x − 8 = −x + 4, so x² − x − 12 = 0 and (x − 4)(x + 3) = 0.

x = 4: line gives y = 0; parabola 16 − 8 − 8 = 0 ✓. x = −3: line gives y = 7; parabola 9 + 6 − 8 = 7 ✓.

(4, 0) and (−3, 7)

Problem 17 Monthly Rate for a Savings Account

A = 1500(1.06^1/12)^12t, and 1.06^1/12 ≈ 1.00487, so the monthly rate is about 0.49%.

After 18 months, t = 1.5: 1500(1.06)^1.5 = 1500(1.09134) ≈ 1637.01.

About 0.49% per month; $1,637.01

Problem 18 Half-Life as Yearly Decay

(12) ^t5 = ((12) ¹⁵ )^t, so b = (12) ¹⁵ ≈ 0.8706.

The yearly decay rate is 1 − 0.8706 = 0.1294, about 12.9%.

At t = 10: 200(12)² = 200 · 14 = 50.

b ≈ 0.8706; about 12.9% per year; 50 grams

Problem 19 Intercepts and End Behavior

f(0) = 3 − 6 = −3, so the y-intercept is (0, −3). Set 3(2)^x − 6 = 0: 2^x = 2, so x = 1, and the x-intercept is (1, 0).

The graph approaches y = −6 on the left. As x → ∞, f(x) → ∞; as x → −∞, f(x) → −6.

(0, −3); (1, 0); asymptote y = −6; → ∞ and → −6

Problem 20 Irrational Intersection Points

x² = 2x + 1, so x² − 2x − 1 = 0. The discriminant is 4 + 4 = 8, so x = 2 ± √82 = 1 ± √2.

x ≈ 2.41 gives y = 2(2.41) + 1 ≈ 5.83. x ≈ −0.41 gives y ≈ 0.17.

x = 1 ± √2; points ≈ (2.41, 5.83) and (−0.41, 0.17)