HOPE Charter School 2009 PSSA Boot Camp
Thursday, January 15, 2009
PSSA Math Reporting Categories: A Numbers and Operations B Measurement C Geometry √ D Algebraic Concepts E Data Analysis and Probability SOLUTION Name (legibly printed): _______________________________________________ Seminar (circle one): Ackerman Beals Capers Harvey Hickson Instructions: • Place your solution into box in front of main office by end of 4th period today. • Winner will be announced during 6th period lunch today. Winner will be randomly selected from all entries, but the answer must be correct and justified (SHOW WORK). • Link with questions and solutions posted at http://hopecharter.blogspot.com
Which equation is graphed below?
A B C D
x+ y =4 2x + y = 8 x + 2y = 8 x − 2y = 8
Solution Method 1: Find the equation of the graphed line and then find which equation matches. The standard form of an equation of a line is y = mx + b where m is the slope of the line and b is the y-intercept.
The y-intercept is where the curve crosses the y-axis: 4. To find the slope, m: ∆y y2 − y1 = m= ∆x x2 − x1 We need to pick two points on the line. It is easiest to pick points where the graphed line crosses the x-axis, and crosses the y-axis (since math with 0’s is easier): ( x1 , y1 ) = (8, 0 ) ← x-intercept
( x2 , y2 ) = ( 0, 4 )
← y-intercept
Substituting: ∆y 4 − 0 4 −1 1 m= = = = =− ∆x 0 − 8 −8 2 2 So, the standard form of an equation of the graphed line is: 1 y =− x+ 4 2 x +y=4
A
−x
−x
y = −x + 4
⇒ m = −1, b = 4 NO
2x + y = 8 B
− 2x
− 2x y = −2 x + 8 ⇒ m = −2, b = 8 NO
x + 2y = 8 −x
−x 2y = − x + 8
C
2 y −x + 8 = 2 2 −x 1 y= + 4 ⇒ m = − , b = 4 YES 2 2
x − 2y = 8 −x
D
−x − 2y = − x + 8 −2 y − x + 8 = −2 −2 x 1 y = − 4 ⇒ m = , b = 4 NO 2 2
Å solution
Solution Method 2:
Since we are provided Multiple Choice (MC) answers of possible linear equations that fit the linear graph, we can simply substitute some points from the graph and see which equation violated or follows the graph. Pick 2 points on the graphed line. Plug the coordinate pair into each equation to see if it passes or fails. It is easiest to pick points where the graphed line crosses the x-axis, and y-axis (since math with 0’s is easier): ( x1 , y1 ) = (8, 0 ) ← x-intercept
( x2 , y2 ) = ( 0, 4 ) A
B
x+ y =4 8+ 0 ≠ 4
← y-intercept
fails with (x1 ,y1 )
2x + y = 8 2 (8) + 0 ≠ 8
fails with (x1 ,y1 )
x + 2y = 8 C
8 + 2 (0) = 8
passes with (x1 ,y1 ) Åsolution
0 + 2 ( 4) = 8
passes with (x 2 ,y 2 )
x − 2y = 8 D
8 − 2 ( 0) = 8
passes with (x1 ,y1 )
0 − 2 ( 4) ≠ 8
fails with (x 2 ,y 2 )