Algebra Placement Exam
Harris School of Public Policy
September 17, 2007
1. Linear Equations. (25 points, 5 each)
Consider the three points P = (3, 2), Q = (4, 7) , and R = (1, 1).
(a) Find the equation of the line passing through points P and Q.
The slop
Algebra Placement Exam
Harris School of Public Policy
August 31, 2007
Solutions
1. Linear Equations. (30 points, 5 each) Consider the line dened by the equation
2x + 5y 7 = 0.
(a) What is the slope of this line?
2
We re-write the equation as 5y = 2x + 7,
Algebra Placement Exam Solutions
Harris School of Public Policy
September 5, 2008
1. Lines. (20 points, 5 each)
(a) Find the equation of the line 1 with slope 2 that passes through the point (7, 4).
3
2
We use Point-Slope Form to nd the equation y 4 = 3 (
Algebra Placement Exam Solutions
Harris School of Public Policy
September 4, 2009
1. Linear Equations I. (20 points, 5 each)
Consider the points A = (6, 4), B = (3, 1) and C = (7, 13) in the Cartesian plane.
(a) Find the equation of the line 1 that passes
Algebra Placement Exam Solutions
Harris School of Public Policy
September 3, 2010
1. Inequalities (21 points, 7 each)
(a) Solve the inequality 1 x + 4 0, and express your answer in interval notation.
2
1
We rst subtract 4 from both sides to get 2 x 4, and
Algebra Placement Exam
Harris School of Public Policy
September 7, 2011
1. Solving Equations (24 points, 6 each) Solve each of the following for the real variable x.
(a)
2
1
=
x
1+x
We rst note that x = 0 and x = 1. But with these conditions in place, we
2012 Calculus Exam Solutions
1.
f (x) =
1
2x
2
+1
x 3x + 1
2
(a)
lim
x2
lim
x2
if x 2
if x > 2
x
+1
2
2
+1=2
2
(b)
3
lim x2 x + 1 = 4 3 + 1 = 2
x2+
2
(c) The function is continuous at x = 2 because it is equal to 2 if you approach it from the left and
the
Algebra Placement Exam
Harris School of Public Policy
September 10, 2012
1. Points and Lines (18 points, 6 each) Let A = (5, 3) and B = (4, 6), and C = (1, 4).
(a) Identify which two of these three points lie on a line with negative slope, and nd
the slop