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Question 75691This question is from textbook Beginning Algebra
: Match the graph with the correct equation on the right.
I duplicated the graph with points (-19, 20)and (20, -19)I am not sure this is correct but the line is showing exactly what the graph is in the book.
The equations are
a)y=-x^2+1
b)y=2x
c)y=x^2-4x
d)y=-x=1
e)y=-x^2+3x
f)y=x^2+1
g)y=x+1
h)y=2x^2
Thanks for your help.
This question is from textbook Beginning Algebra
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! The most important thing in this problem is to learn major characteristics of equations ...
characteristics that will help you to get a sense of what the graph looks like
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a)y=-x^2+1
This graph is a parabola because it has x-squared as its highest exponential term. The fact
that the x-squared term has a minus sign means the the parabola rises to a peak and then
falls back down. Think of the path of a football pass, a basketball shot, or a volleyball
serve. To find where the graph crosses the x-axis, set y equal to zero and solve.
You will get:
.
0 = -x^2 + 1
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Add x^2 to both sides in order to eliminate the minus x^2 on the right side. The equation
then becomes x^2 = 1 and when you solve for x by taking the square root of both sides, you
find that the graph has x = 1 and x = -1 as the two points where the graph crosses
the x axis ... all because you made y equal to zero. The peak of the graph will occur
midway between those to points, and if you sketch the coordinate system you will see that
the y-axis is midway between x = -1 and x = +1. At that point x must be zero. Return to
the original equation and set x equal to zero. You will get y = +1. Mark +1 on the y-axis
and you now have three points on the graph ... (-1,0), (+1,0), and (0,+1). That should give
you a pretty good idea of what the graph looks like.
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b)y=2x
.
No x^2 term and x only has an exponent of 1. These are characteristics of an equation
having a straight line as a graph. The equation is of the slope-intercept form:
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y = mx + b
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in which m is the slope of the equation and b is the point where the graph crosses
the y-axis. Since there is no "b" term in y = 2x, then b = 0. This tells you that the
graph goes across the y-axis where y is equal to zero. [You could also do this by setting
x equal to 0 in the equation y = 2x and when you do you find that y = 0. So a point on
the graph is (0,0).] And from the slope-intercept form you can tell that the slope
(the multiplier of the x) is +2. Since it is positive, you know that the graph goes up
and to the right. The rate at which it slants is up two vertical units for every
one unit you move horizontally to the right. You know that the origin (0,0) is on the
graph. From this point move 1 unit to the right along the x-axis. Then move 2 units
vertically up. That should put you at the point (+1, +2) and it is on the graph.
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c)y=x^2-4x
.
The analysis of this is similar to that of problem a) above. It is a parabola. The x^2
term is positive and therefore the graph is "bowl" shaped ... it starts high to the left,
falls to a low point, and then turns upward again as you look to the right side of the
graph. An easy point to find is what happens when x equals zero. If you substitute
zero for x in the equation, you find that y then also equals zero. From this you know
that (0,0) is on the graph. Next set y equal to zero. This will tell you where the graph
crosses the x-axis. When you set y = 0 the equation becomes:
.
0 = x^2 - 4x
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Factor this equation and you get:
.
0 = x*(x - 4)
.
This equation will be true if either of the two factors is zero. [If either factor is
zero, the right side has a multiplication by zero and therefore equals the left side
of the equation.] Set the factors equal to zero and solve for x. When you do, you get
x = 0 or x = +4. So the graph crosses the x-axis where x equals 0 and also where
x equals +4. The low point of this parabola will occur midway between 0 and 4 on the
x-axis ... in other words, the low point on the graph occurs when x equals 2. Plugging
2 in for x in the equation results in y = (2)^2 - 4*2 = 4 - 8 = -4. So the low point on
the graph is at (2, -4)
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d)y=-x+1
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Same analysis as problem b) above. This time the graph crosses the y-axis at +1 (the value
of b). Also the slope is -1 which, since it is negative, means that the graph slopes
downward as you move to the right. The rate of the slope is 1, meaning that for every
unit that you move horizontally to the right, you have to move 1 unit vertically downward
to get back to the graph. You know that (0,+1) is a point on the graph because it
contains the coordinates of the y-axis crossing. Move 1 unit horizontally to the right
from that point and then move 1 unit vertically down. That should put you at the point
(1,0) and that point should be on the graph. You can also get this point by setting
y equal to zero in the equation and solving for x. That method should also give you the
point (1, 0).
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e)y=-x^2+3x
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By now you should be able to tell that this is a parabola that is a lot like the one in
problem a) above. It has a minus sign on the x^2 term meaning that the curve rises to
a peak and then falls back down as you move to the right. How can you tell the difference
though between this graph and the one in problem a)? How about setting y = to zero
and then solving for the two values of x. If you follow the same process as you did in
problem a) and problem c) you will find that when y equals zero the values of x that come
out by factoring that x = 0 is one value and x = +3 is the other. So the graph has
(0,0) and (3,0) as two points on the graph. You now know that the peak occurs where x
is midway between 0 and 3 on the x-axis. This midway point is x = 1.5 and you can,
if you want to, substitute 1.5 in for x in the equation and find the corresponding
value of y at the peak.
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f)y=x^2+1
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This is a lot like problem a) except this time the x^2 term has positive sign. This tells
you that the parabola is "bowl-shaped" in that it falls to a low point and then rises back
upward as you move to the right. One thing you can do easily is to set x equal to zero
and you find that the corresponding value of y is +1. So you know that (0,1) is on the
graph.
If you set y = 0 and then try to solve for x you run into a problem because you get
x^2 = -1. But you can't solve this by taking the square root of both sides because
there is no real number that can be multiplied by itself to give -1. This tells you that
the graph does not cross or touch the x-axis.
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g)y=x+1
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Has to be a straight-line graph because x terms are only first order (exponent of 1).
This is in the slope intercept form. The slope (multiplier of x) is +1 so the graph
goes up and to the right (positive slope). It crosses the y-axis at +1. You can set
y = 0 and solve for x to find that the graph crosses the x-axis at x = -1.
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h)y=2x^2
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You can tell from this equation that when y is set equal to zero, then x must also equal
zero. So the point (0,0) is on the graph. Also when x is set equal to zero, then
y must also equal zero. Therefore, (0,0) is the low point on the graph. You can also
tell that the graph is a parabola (because the x is squared) and that the graph is bowl
shaped opening upward because the sign on the x-squared term is positive. You could also
let x = 1 and solve for y to find that y will equal 2 when x equals 1. So (1,2) is a
point on the graph. Also by letting x = -1, you get that y equals 2. So (-1,2) is also
a point on the graph. If you think about it, this tells you that the y-axis is the
"centerline" of this graph.
.
Hope the above explanations help you to think about the characteristics of graphs of
parabolic and straight-line equations. And I hope this is what you were looking for in
terms of how to match the given equations to graphs given in the book.
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