Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Cumulative Standards Review

Exercise 12 Page 159

Find a point on the graph each line passes through, and see if it satisfies the given equation.

G

Practice makes perfect
We need to find which graph correctly displays the given equation. t=4c Observing each graph, we can find a lattice point through which the lines pass. Then, test to see if those points satisfy the equation. If any point does not satisfy the equation, then the graph cannot correctly display the relationship.
Graph Point (c,t) Substitute Simplify Satisfies the Equation?
F ( 2, 9) 9? =4* 2 9≠ 8 No
G ( 3, 12) 12? =4* 3 12=12 Yes
H ( 4, 12) 12? =4* 4 12≠ 16 No
I ( 1, 3) 3? =4* 1 3≠ 4 No

Solutions F, H, and I have points that do not satisfy the equation. For a graph to correctly display a relationship, all points on the graph must satisfy the given equation. Having even one point that does not satisfy the equation immediately disqualifies the graph – leaving us with option G.

Double Check Our Answer

To double check our answer, we can find another point on graph G and check to see if it also satisfies the equation. We can see that the graph passes through the origin, (0,0), so let's check that point.
t? =4c
0? =4* 0
0=0
Graph G correctly displays the relation t=4c.

Alternative Solution

Using the origin
The graph of any equation that has no constant passes through the origin (0,0). We can test the given equation t=4c to see if it passes through the origin by substituting the point (0,0) for t and c.
t? =4c
0? =4* 0
0=0
The graph of t=4c passes through the origin. This means that right off the bat we can disregard options F and H because they do not pass through the origin. Let's examine the remaining graphs.
We see that when c=3, t is an integer for both G and I. By substituting c=3 into the equation, we can find which t value matches our equation.
t=4c
t=4* 3
t=12
When c=3, t=12. This result corresponds to option G.