MATH 110 MIDTERM SOLUTIONS!
Here are the solutions to your exam. Both version’s solutions are here. Enjoy!!!
Midterm A solutions
p
√
(1) (a) f (−1) = 1 + (−1)2 = 2.
(b) The y−intercepts occur when x = 0,p
so this is only √
at y = 1.
(c) The function is even, since f (−x) = 1 + (−x)2 = 1 + x2 = f (x).
(2) The graph would be. . .
(a) shifted to the left 4 units.
(b) compressed toward the y−axis by a factor of 3.
(c) stretched vertically (away from the x−axis) by a factor of 3, then shifted up
1 unit.
p
√
√
(3) The distance d = (7 − (−1))2 + (2 − 5)2 = 64 + 9 = 73.
(4) Completing the square on x2 + 6x, we see that b = 6, b/2 = 3, so (b/2)2 = 9, and so
x2 + 6x = x2 + 6x + 9 − 9 = (x + 3)2 − 9. Doing the same process with y 2 − 10y, we
have b = −10, b/2 = −5, and (b/2)2 = 25, so that y 2 − 10y = y 2 − 10y + 25 − 25 =
(y − 5)2 − 25. Plugging these values into the original equation, we have
2 = x2 + 6x + y 2 − 10y = (x + 3)2 − 9 + (y − 5)2 − 25, so 36 = (x + 3)2 + (y − 5)2 .
√
Thus, the center is (−3, 5), and the radius is 36 = 6.
(5) The slope of this line is the same as the one given, which is 3. Using the point
slope form of a line, we have the equation as y − 2 = 3(x − 8), or y = 3x − 22.
(6) The slope of this line is the negative reciprocal of the one given, so its slope would
be −1
4 . Again using the point slope form of a line, we have our equation as y − 5 =
−1
(x
− 4), or y = −1
4
4 x + 6.
√
3(2/3)−2
(7) (a) f ( 23 ) = (2/3)2 −2(2/3)+1 = 0, since the numerator of this is zero and the denominator isn’t.
(b) The inputs of this function can’t make 3x − 2 < 0, which is when 3x < 2, or
x < 32 . Also, the denominator x2 − 2x + 1 = (x − 1)2 cannot be zero, which occurs
only at x = 1. So the domain is all real x with x ≥ 23 , and x 6= 1.
1
2
MATH 110 MIDTERM SOLUTIONS!
Midterm B solutions
(1) (a) f (−1) = (−1)3 − (−1) = −1 + 1 = 0.
(b) The y−intercepts occur when x = 0, so this is only at y = 0.
(c) The function is odd, since f (−x) = (−x)3 − (−x) = −x3 + x = −(x3 − x) =
−f (x).
(2) The graph would be. . .
(a) shifted to the right 3 units.
(b) stretched vertically (away from the x−axis) by a factor of 2.
(c) compressed p
toward the y−axis by a√factor of 5,
√ then shifted up 1 unit.
2
2
(3) The distance d = (3 − 0) + (2 − 3) = 9 + 1 = 10.
(4) Completing the square on x2 + 6x, we see that b = 6, b/2 = 3, so (b/2)2 = 9, and so
x2 +6x = x2 +6x+9−9 = (x+3)2 −9. Doing the same process with y 2 −4y, we have
b = −4, b/2 = −2, and (b/2)2 = 4, so that y 2 − 4y = y 2 − 4y + 4 − 4 = (y − 2)2 − 4.
Plugging these values into the original equation, we have
−4 = x2 + 6x + y 2 − 4y = (x + 3)2 − 9 + (y − 2)2 − 4, so 9 = (x + 3)2 + (y − 2)2 .
√
Thus, the center is (−3, 2), and the radius is 9 = 3.
(5) The slope of this line is the same as the one given, which is 1. Using the point
slope form of a line, we have the equation as y − 3 = 1(x − 5), or y = x − 2.
(6) The slope of this line is the negative reciprocal of the one given, so its slope would
be −1
2 . Again using the point slope form of a line, we have our equation as y − 1 =
−1
3
(x
− 1), or y = −1
2
2 x + 2.
√
2(1/2)−1
(7) (a) f ( 12 ) = (1/2)2 −6(1/2)+9 = 0, since the numerator of this is zero and the denominator isn’t.
(b) The inputs of this function can’t make 2x − 1 < 0, which is when 2x < 1, or
x < 21 . Also, the denominator x2 − 6x + 9 = (x − 3)2 cannot be zero, which occurs
only at x = 3. So the domain is all real x with x ≥ 12 , and x 6= 3.