Difference between revisions of "1992 AHSME Problems/Problem 14"

(Created page with "== Problem == Which of the following equations have the same graph? <math>I.\quad y=x-2 \qquad II.\quad y=\frac{x^2-4}{x+2}\qquad III.\quad (x+2)y=x^2-4</math> <math>\text{(A)...")
 
(Solution: added solution)
 
Line 12: Line 12:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
+
The equations differ at <math>x=-2</math>. The graph of <math>I</math> would contain the point <math>(-2, -4)</math>; <math>-2</math> is undefined in the graph of <math>II</math> because it gives a denominator of <math>0</math>; and the graph of <math>III</math> contains the whole the vertical line <math>x = -2</math> as for any <math>y</math> value, the equation still satisfies <math>0 = 0</math>. So the answer is <math>{E}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:45, 18 January 2018

Problem

Which of the following equations have the same graph?

$I.\quad y=x-2 \qquad II.\quad y=\frac{x^2-4}{x+2}\qquad III.\quad (x+2)y=x^2-4$

$\text{(A) I and II only} \quad \text{(B) I and III only} \quad \text{(C) II and III only} \quad \text{(D) I,II,and III} \quad \\ \text{(E) None. All of the equations have different graphs}$

Solution

The equations differ at $x=-2$. The graph of $I$ would contain the point $(-2, -4)$; $-2$ is undefined in the graph of $II$ because it gives a denominator of $0$; and the graph of $III$ contains the whole the vertical line $x = -2$ as for any $y$ value, the equation still satisfies $0 = 0$. So the answer is ${E}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png