This algebra 1 graphs worksheet answer key solves all 23 problems of the graphs as solution sets, linear-quadratic systems and exponential expressions worksheet step by step.
Each worked solution shows the substitution or the factoring, the check in both equations, and the final answer in bold. If your intersection points differ, recheck the signs when you move terms to one side and the order of the coordinates.
Algebra 1 Graphs Worksheet Answer Key: problems 1–5
Problem 1: Is the point a solution?
(a) \(2(3)-1=5\), which equals the y-value 5. Yes, it is a solution.
(b) \(2(1)-1=1\), but the y-value is 2. No, it is not a solution, so \((1,2)\) is not on the graph.
Problem 2: Points on a parabola
\((-2)^{2}+3=4+3=7\), so \((-2,7)\) is on the graph.
\(3^{2}+3=12\), not 10, so \((3,10)\) is not on the graph.
Problem 3: Find the missing coordinate
\(k=4^{2}-6=10\).
\(3=m^{2}-6\) gives \(m^{2}=9\), so \(m=3\) or \(m=-3\). Check: \((-3)^{2}-6=3\). \(k=10\); \(m=3\) or \(m=-3\)
Problem 4: Intercepts as solutions
With \(y=0\): \(0=2x-6\), so \(x=3\). With \(x=0\): \(y=-6\). x-intercept \((3,0)\), y-intercept \((0,-6)\)
Problem 5: Points and a graph
A: \(-1+4=3\) yes. B: \(-3+4=1\neq 2\) no. C: \(-0+4=4\) yes. D: \(-(-2)+4=6\) yes. E: \(-4+4=0\neq 1\) no.
A, C and D are solutions; they are the points on the line.
Algebra 1 Graphs Worksheet Answer Key: problems 6–10
Problem 6: Line and parabola: y = x² and y = x + 6
\(x^{2}=x+6\) gives \(x^{2}-x-6=0\), so \((x-3)(x+2)=0\) and \(x=3\) or \(x=-2\).
\(y=3+6=9\) and \(y=-2+6=4\). Check: \(3^{2}=9\) and \((-2)^{2}=4\). \((3,9)\) and \((-2,4)\)
Problem 7: System with a trinomial
\(x^{2}+2x-3=x+3\) gives \(x^{2}+x-6=0\), so \((x+3)(x-2)=0\) and \(x=-3\) or \(x=2\).
\(y=-3+3=0\) and \(y=2+3=5\). Check: \(9-6-3=0\) and \(4+4-3=5\). \((-3,0)\) and \((2,5)\)
Problem 8: Read and verify
The graphs meet at \((0,-3)\) and \((3,0)\).
Algebra: \(x^{2}-2x-3=x-3\) gives \(x^{2}-3x=0\), so \(x(x-3)=0\) and \(x=0\) or \(x=3\). Then \(y=-3\) and \(y=0\). \((0,-3)\) and \((3,0)\)
Problem 9: How many solutions?
(a) \(x^{2}-4=5\) gives \(x^{2}=9\), so \(x=\pm 3\): 2 solutions, \((3,5)\) and \((-3,5)\).
(b) \(x^{2}-4=-4\) gives \(x^{2}=0\): 1 solution, \((0,-4)\).
(c) \(x^{2}-4=-6\) gives \(x^{2}=-2\), which has no real solution: 0 solutions. 2, 1 and 0 solutions
Problem 10: A line that touches the parabola
\(x^{2}+2=2x+1\) gives \(x^{2}-2x+1=0\), so \((x-1)^{2}=0\) and \(x=1\) only.
\(y=2(1)+1=3\). Check: \(1+2=3\). One solution, \((1,3)\)
Algebra 1 Graphs Worksheet Answer Key: problems 11–15
Problem 11: Rearranging before factoring
\(x^{2}-3x=2x-6\) gives \(x^{2}-5x+6=0\), so \((x-2)(x-3)=0\) and \(x=2\) or \(x=3\).
\(y=2(2)-6=-2\) and \(y=2(3)-6=0\). Check: \(4-6=-2\) and \(9-9=0\). \((2,-2)\) and \((3,0)\)
Problem 12: Downward parabola
\(-x^{2}+4=-x+2\) gives \(0=x^{2}-x-2\), so \((x-2)(x+1)=0\) and \(x=2\) or \(x=-1\).
\(y=-2+2=0\) and \(y=1+2=3\). Check: \(-4+4=0\) and \(-1+4=3\). \((2,0)\) and \((-1,3)\)
Problem 13: Table for a growth function
\(4^{-2}=\frac{1}{16}\), \(4^{-1}=\frac{1}{4}\), \(4^{0}=1\), \(4^{1}=4\), \(4^{2}=16\), \(4^{3}=64\).
The graph crosses the y-axis at \((0,1)\). As \(x\) decreases, the outputs shrink toward 0 but stay positive, so the graph approaches the horizontal line \(y=0\).
Problem 14: Features of a growth function
At \(x=0\), \(y=5\cdot 1=5\), so the y-intercept is \((0,5)\). Since \(5\cdot 3^{x}>0\) for every \(x\), there is no x-intercept.
As \(x\) gets very large, \(y\) gets very large (\(y=15\) at \(x=1\), \(y=45\) at \(x=2\)). As \(x\) gets very small, \(y\) approaches 0 (\(y=\frac{5}{3}\) at \(x=-1\)).
Problem 15: Exponential decay
The y-intercept is \((0,64)\). As \(x\) gets large, \(y\) approaches 0; as \(x\) gets very negative, \(y\) grows without bound.
\(y(5)=64\cdot\frac{1}{32}=2\). y-intercept \((0,64)\); \(y=2\)
Algebra 1 Graphs Worksheet Answer Key: problems 16–20
Problem 16: A shifted exponential
y-intercept: \(2^{0}-4=-3\), so \((0,-3)\). x-intercept: \(2^{x}-4=0\) gives \(2^{x}=4\), so \(x=2\) and the point is \((2,0)\). The graph approaches \(y=-4\) on the left.
\((0,-3)\), \((2,0)\), asymptote \(y=-4\)
Problem 17: Rewrite with base 2
(a) \(8^{x}=(2^{3})^{x}=2^{3x}\).
(b) \(4^{3x}=(2^{2})^{3x}=2^{6x}\).
(c) \(16^{x/2}=(2^{4})^{x/2}=2^{2x}\). \(2^{3x}\), \(2^{6x}\), \(2^{2x}\)
Problem 18: A two-year factor
\(1.2^{t}=(1.2^{2})^{t/2}=1.44^{t/2}\), so \(P(t)=500(1.44)^{t/2}\). Check: \(500(1.2)^{2}=500(1.44)=720\) and \(500(1.44)^{1}=720\). \(P(t)=500(1.44)^{t/2}\); the 2-week factor is 1.44, a 44% increase
Problem 19: Ball on a hill
\(-x^{2}+6x=x\) gives \(-x^{2}+5x=0\), so \(x(5-x)=0\) and \(x=0\) (the kick) or \(x=5\).
\(y=5\). Check: \(-25+30=5\). The ball lands at \((5,5)\), 5 feet along and 5 feet up
Problem 20: Savings account, monthly rate
\(1.06^{1/12}\approx 1.0049\), so \(A=(1.0049)^{12t}\). The monthly rate is about 0.49%.
Check: \(1.0049^{12}\approx 1.0603\), close to 1.06 (rounding).
Algebra 1 Graphs Worksheet Answer Key: problems 21–23
Problem 21: Growth of 15% per year
\(1.15^{t}=(1.15^{1/12})^{12t}\approx(1.0117)^{12t}\), so \(N\approx 800(1.0117)^{12t}\). The monthly rate is about 1.17%.
This is less than \(15\%: 12=1.25\%\) because of compounding. Check: \(1.0117^{12}\approx 1.1496\), about 1.15.
Problem 22: Car value each quarter
\(0.85^{t}=(0.85^{1/4})^{4t}\) and \(0.85^{1/4}\approx 0.9602\), so \(V\approx 22000(0.9602)^{4t}\). The factor 0.9602 means the value drops by \(1-0.9602=0.0398\), about 4.0% per quarter.
Check: \(0.9602^{4}\approx 0.85\).
Problem 23: Bacteria every 20 minutes
\(2^{t}=(2^{1/3})^{3t}\) and \(2^{1/3}\approx 1.26\), so \(N\approx 100(1.26)^{3t}\). The culture grows about 26% every 20 minutes.
Check: \(1.26^{3}\approx 2.0\).
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