Download Solutions - UCR Math Dept.

Josh Buli
MATH 008A - Quiz 2 Solutions
2014
1) Find the x and y intercepts: x2 + 2y 2 = 2
Solution: To find the x-intercepts, plug in zero for y and solve for x. To find the y-intercepts, plug in zero
for x and solve for y.
x2 + 2y 2 = 2
x2 = 2
√
x=± 2
and
x2 + 2y 2 = 2
2y 2 = 2
y = ±1
√
So the coordinates are (± 2, 0) and (0, ±1).
2) Find the equation of the line passing through (3, −2) that is perpendicular to 3x + 2y = 2.
Solution: Putting the equation for a line that is given into slope intercept form, we have
3
y =− x+1
2
So we have that the slope of this line is m = − 32 . To find the slope of the perpendicular line, take the
negative reciprocal, and we have that m⊥ = 23 . So now using the point slope formula, we have that
y − y1 = m⊥ (x − x1 )
2
y − (−2) = (x − 3)
3
2
y+2= x−2
3
2
y = x−4
3
3) Identify the center and the radius : x2 + y 2 − 6x + 4y − 3 = 0
Solution: First group the x and y terms together and complete the square
(x2 − 6x) + (y 2 + 4y) = 3
(x2 − 6x + 9) + (y 2 + 4y + 4) = 3 + 9 + 4
(x − 3)2 + (y + 2)2 = 16
So we have that the center is (3, −2) and the radius is 4.
1
2
f (x + h) − f (x)
for the function”
h
f (x) = −2x2 − 3x + 5
4) Find and simplify the difference quotient
Solution: Remember composition of functions to find f (x + h), then
−2(x + h)2 − 3(x + h) + 5 − (−2x2 − 3x + 5)
f (x + h) − f (x)
=
h
h
−2x2 − 4xh − 2h2 − 3x − 3h + 5 + 2x2 + 3x − 5
=
h
−4xh − 2h2 − 3h
=
h
− 6 h(4x + 2h + 3)
=
6h
= −(4x + 2h + 3)
5) Given f (x) = 2x−1
x+3 , find the domain of f and the x and y intercepts.
Solution: Remember that the domain of a function is the places where the function is defined. The only
place where f is not defined is at x = −3, so the domain is (−∞, −3) ∪ (−3, ∞).
The x and y intercepts are found the same way as problem 1.
2x − 1
y=
x+3
2(0) − 1
y=
0+3
1
y=−
3
and
2x − 1
y=
x+3
2x − 1
0=
x+3
0 = 2x − 1
1
x=
2
So we have that the x-intercept is ( 12 , 0) and (0, − 31 ).