MATH1051: Exam 3
Vishal Saraswat
Here is the key to the third exam. This exam covered chapters 3 and 4 from the book. Be sure you understand how
to do each problem since you will see similar problems on the final exam.
1. (10 points) Find the equation of the parabola that has vertex at (2, −3) and y-intercept at (0, −1). Write your
answer in standard form: (y = ax2 + bx + c).
Solution:
Use the vertex form of the equation of a parabola:
y = a(x − h)2 + k
where (h, k) is the vertex of the parabola.
So we can write the required equation as
y
15
y = a(x − 2)2 + (−3) .
We are also given the point (0, −1) on the
parabola so we can use those values of x
and y to find a.
f (x) = 12 x2 − 2x − 1
10
y = a(x − 2)2 − 3
or, − 1 = a(0 − 2)2 − 3
or, − 1 = a(4) − 3
or, − 1 + 3 = 4a
2
or, = a .
4
5
−x
−10
−5
O (0, −1)
So we can write the required equation as
1
y = (x − 2)2 − 3
2
5
x
10
(2, −3)
−5
−y
which in standard form is
1
1
1
y = (x − 2)2 − 3 = (x2 − 4x + 4) − 3 = x2 − 2x − 1 .
2
2
2
1
Ans: y = x2 − 2x − 1
2
You can also do this problem by just using the standard form of the equation by using the fact
that the points (2, −3) and (0, −1) lie on the parabola and that x-coordinate of the vertex is
b
given by − 2a
.
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 3
Vishal Saraswat
2. Given the function f (x) = −2x2 + 4x − 6 find the following:
(a) (5 points) x-intercept(s), if any.
The x-intercept is the point at which a curve or function crosses the x-axis, that is,
when f (x) = 0. Now,
Solution:
y
f (x) = 0
2
iff − 2x + 4x − 6 = 0
iff x2 − 2x + 3 = 0
−x
x
−5
iff x2 − 2x + 1 + 2 = 0
iff (x − 1)2 + 2 = 0
2
iff (x − 1) = −2 .
Since (x−1)2 cannot be negative, there is no solution
to f (x) = 0. Thus there is no x-intercept of the
function f (x) = −2x2 + 4x − 6.
O
5
(1, −4)
−5
(0, −6)
−10
Ans: None
You can also use the quadratic formula to deduce
that there is no solution to the equation f (x) = 0.
Note that the graph of the function f (x) is a parabola
pointing upwards with vertex at (1, −4) and the
graph of the parabola is always below the vertex.
So, the graph never crosses the x-axis.
−15
−20
−yf (x) = −2x2 + 4x − 6
(b) (5 points) y-intercept(s), if any.
The y-intercept is the point at which a curve or function crosses the y-axis, that is,
the point f (0). Now,
f (0) = −2 × 02 + 4 × 0 − 6 = −6 .
Solution:
Thus the y-intercept is (0, −6).
Ans: −6
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 3
Vishal Saraswat
3. Given the function f (x) = −2x2 + 4x − 6 find the following:
(a) (5 points) The coordinates of the vertex.
The x-coordinate of the vertex of the parabola defined by the function
. Here, a = −2
f (x) = ax + bx + c is given by −b
2a
y
and b = 4. Thus, the x-coordinate of the vertex of
the parabola is given by
−x
Solution:
2
−b
−4
−4
=
=
= 1.
2a
2 × −2
−4
−5
Now,
f (1) = −2 × 12 + 4 × 1 − 6 = −2 + 4 − 6 = −4 .
x
5
O
(1, −4)
−5
(0, −6)
So the coordinates of the vertex are (1, −4).
Ans: (1, −4)
−10
−15
−20
−yf (x) = −2x2 + 4x − 6
(b) (5 points) Equation of the axis of symmetry.
The axis of symmetry of the parabola defined by the function f (x) = ax2 + bx + c
is given by x = −b
. Thus the axis of symmetry of the parabola defined by the function f (x) =
2a
−2x2 + 4x − 6 is
−4
−4
x=
=
= 1.
2 × −2
−4
Solution:
Ans: x = 1
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 3
Vishal Saraswat
4. Given the function f (x) = (x + 3)3 (x − 1)(x + 1)2 determine:
(a) (5 points) where the graph CROSSES the x-axis, if it does.
Solution:
Ans: x = −3 and x = 1
(b) (5 points) where the graph TOUCHES the x-axis, if it does.
Solution:
Ans: x = −1
(c) (5 points) the end behavior of the graph (i.e., what power function does the graph look like for large x?).
Solution:
Ans: x6
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 3
Vishal Saraswat
5. (10 points) Solve algebraically: 6 − x ≤ x2 . You must be precise with your final answer.
Solution:
6 − x ≤ x2
iff 0 ≤ x2 + x − 6
iff 0 ≤ (x + 3)(x − 2)
iff either x ≤ −3 or x ≥ 2 .
Ans: x ≤ −3; x ≥ 2
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 3
Vishal Saraswat
6. (15 points) Solve algebraically:
x3 + x − 2
> x. You must be precise with your final answer.
x2 − x
Solution:
iff
iff
iff
iff
iff
iff
x3 + x − 2
>x
x2 − x
x3 + x − 2
−x>0
x2 − x
x3 + x − 2 − x(x2 − x)
>0
x2 − x
x3 + x − 2 − x3 + x2
>0
x2 − x
x2 + x − 2
>0
x2 − x
(x + 2)(x − 1)
>0
x(x − 1)
either x < −2 or 0 < x < 1 or x > 1 .
Ans: x < −2; 0 < x < 1; x > 1
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 6 of ??
MATH1051: Exam 3
7. Given the function: f (x) =
Vishal Saraswat
x3 − 3x2 + 2x
, use the space below to find the important parts of the graph such as
x2 − x − 2
(a) (3 points) intercepts,
There are two x-intercepts: at x = 0 and x = 1.
The y-intercept is at y = 0.
Solution:
(b) (10 points) asymptotes,
Solution:
There is only one vertical assymptote:
x = −1.
Some can get a wrong one too:
x=2
which is not a vertical assymptote but that’s where there is a hole, that is, there exists a limit
to the function at x = 2.
There is no horizontal assymptote and there is one oblique assymptote:
y = x − 2.
(c) (4 points) intersections with the asymptotes,
Solution:
Ans: None
(d) (3 points) and holes.
The function is not defined at at x = 2 but the limit at x = 2 exists and equal to
2/3. So there is a hole at (2, 2/3).
Solution:
Ans:
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 7 of ??
MATH1051: Exam 3
Vishal Saraswat
3
2
+2x
8. (10 points) Graph the function: f (x) = x x−3x
using the information from the previous question. Be precise
2 −x−2
when plotting points and drawing asymptotes.
Solution:
y
x = −1
7
6
f (x) =
x3 −3x2 +2x
x2 −x−2
y =x−2
5
4
3
2
1
−x
−7
−6
−5
−4
−3
−2
−1
O
(2, 2/3)
1
2
3
4
5
6
x
7
−1
−2
−3
−4
−5
−6
−7
−8
−9
f (x) =
x3 −3x2 +2x
x2 −x−2
−10
−y
Ans:
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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