document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=75691>Question 75691</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0072549017/algebrahomeworkh>Beginning Algebra </A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Match the graph with the correct equation on the right. <BR>\n" );
document.write( "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.<BR>\n" );
document.write( "The equations are <BR>\n" );
document.write( "a)y=-x^2+1<BR>\n" );
document.write( "b)y=2x<BR>\n" );
document.write( "c)y=x^2-4x<BR>\n" );
document.write( "d)y=-x=1<BR>\n" );
document.write( "e)y=-x^2+3x<BR>\n" );
document.write( "f)y=x^2+1<BR>\n" );
document.write( "g)y=x+1<BR>\n" );
document.write( "h)y=2x^2\r<BR>\n" );
document.write( "\n" );
document.write( "Thanks for your help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #54371 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=54371\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The most important thing in this problem is to learn major characteristics of equations ...<BR>\n" );
document.write( "characteristics that will help you to get a sense of what the graph looks like <BR>\n" );
document.write( ".<BR>\n" );
document.write( "a)y=-x^2+1<BR>\n" );
document.write( "This graph is a parabola because it has x-squared as its highest exponential term.  The fact<BR>\n" );
document.write( "that the x-squared term has a minus sign means the the parabola rises to a peak and then<BR>\n" );
document.write( "falls back down. Think of the path of a football pass, a basketball shot, or a volleyball <BR>\n" );
document.write( "serve.  To find where the graph crosses the x-axis, set y equal to zero and solve.  <BR>\n" );
document.write( "You will get:<BR>\n" );
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document.write( "0 = -x^2 + 1<BR>\n" );
document.write( ".<BR>\n" );
document.write( "Add x^2 to both sides in order to eliminate the minus x^2 on the right side. The equation<BR>\n" );
document.write( "then becomes x^2 = 1 and when you solve for x by taking the square root of both sides, you<BR>\n" );
document.write( "find that the graph has x = 1 and x = -1 as the two points where the graph crosses<BR>\n" );
document.write( "the x axis ... all because you made y equal to zero. The peak of the graph will occur<BR>\n" );
document.write( "midway between those to points, and if you sketch the coordinate system you will see that<BR>\n" );
document.write( "the y-axis is midway between x = -1 and x = +1.  At that point x must be zero. Return to<BR>\n" );
document.write( "the original equation and set x equal to zero.  You will get y = +1. Mark +1 on the y-axis<BR>\n" );
document.write( "and you now have three points on the graph ... (-1,0), (+1,0), and (0,+1). That should give<BR>\n" );
document.write( "you a pretty good idea of what the graph looks like.<BR>\n" );
document.write( ".<BR>\n" );
document.write( "b)y=2x<BR>\n" );
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document.write( "No x^2 term and x only has an exponent of 1.  These are characteristics of an equation<BR>\n" );
document.write( "having a straight line as a graph.  The equation is of the slope-intercept form:<BR>\n" );
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document.write( "y = mx + b<BR>\n" );
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document.write( "in which m is the slope of the equation and b is the point where the graph crosses <BR>\n" );
document.write( "the y-axis.  Since there is no \"b\" term in y = 2x, then b = 0.  This tells you that the<BR>\n" );
document.write( "graph goes across the y-axis where y is equal to zero.  [You could also do this by setting<BR>\n" );
document.write( "x equal to 0 in the equation y = 2x and when you do you find that y = 0. So a point on<BR>\n" );
document.write( "the graph is (0,0).] And from the slope-intercept form you can tell that the slope <BR>\n" );
document.write( "(the multiplier of the x) is +2. Since it is positive, you know that the graph goes up<BR>\n" );
document.write( "and to the right.  The rate at which it slants is up two vertical units for every <BR>\n" );
document.write( "one unit you move horizontally to the right.  You know that the origin (0,0) is on the<BR>\n" );
document.write( "graph. From this point move 1 unit to the right along the x-axis. Then move 2 units<BR>\n" );
document.write( "vertically up. That should put you at the point (+1, +2) and it is on the graph.<BR>\n" );
document.write( ". <BR>\n" );
document.write( "c)y=x^2-4x<BR>\n" );
document.write( ".<BR>\n" );
document.write( "The analysis of this is similar to that of problem a) above.  It is a parabola. The x^2<BR>\n" );
document.write( "term is positive and therefore the graph is \"bowl\" shaped ... it starts high to the left, <BR>\n" );
document.write( "falls to a low point, and then turns upward again as you look to the right side of the<BR>\n" );
document.write( "graph. An easy point to find is what happens when x equals zero.  If you substitute<BR>\n" );
document.write( "zero for x in the equation, you find that y then also equals zero.  From this you know<BR>\n" );
document.write( "that (0,0) is on the graph. Next set y equal to zero.  This will tell you where the graph<BR>\n" );
document.write( "crosses the x-axis. When you set y = 0 the equation becomes:<BR>\n" );
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document.write( "0 = x^2 - 4x<BR>\n" );
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document.write( "Factor this equation and you get:<BR>\n" );
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document.write( "0 = x*(x - 4)<BR>\n" );
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document.write( "This equation will be true if either of the two factors is zero. [If either factor is<BR>\n" );
document.write( "zero, the right side has a multiplication by zero and therefore equals the left side<BR>\n" );
document.write( "of the equation.] Set the factors equal to zero and solve for x.  When you do, you get<BR>\n" );
document.write( "x = 0 or x = +4.  So the graph crosses the x-axis where x equals 0 and also where <BR>\n" );
document.write( "x equals +4.  The low point of this parabola will occur midway between 0 and 4 on the<BR>\n" );
document.write( "x-axis ... in other words, the low point on the graph occurs when x equals 2.  Plugging<BR>\n" );
document.write( "2 in for x in the equation results in y = (2)^2 - 4*2 = 4 - 8 = -4. So the low point on<BR>\n" );
document.write( "the graph is at (2, -4)<BR>\n" );
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document.write( "d)y=-x+1<BR>\n" );
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document.write( "Same analysis as problem b) above.  This time the graph crosses the y-axis at +1  (the value<BR>\n" );
document.write( "of b). Also the slope is -1 which, since it is negative, means that the graph slopes<BR>\n" );
document.write( "downward as you move to the right.  The rate of the slope is 1, meaning that for every<BR>\n" );
document.write( "unit that you move horizontally to the right, you have to move 1 unit vertically downward<BR>\n" );
document.write( "to get back to the graph.  You know that (0,+1) is a point on the graph because it <BR>\n" );
document.write( "contains the coordinates of the y-axis crossing.  Move 1 unit horizontally to the right <BR>\n" );
document.write( "from that point and then move 1 unit vertically down.  That should put you at the point <BR>\n" );
document.write( "(1,0) and that point should be on the graph.  You can also get this point by setting<BR>\n" );
document.write( "y equal to zero in the equation and solving for x.  That method should also give you the<BR>\n" );
document.write( "point (1, 0).<BR>\n" );
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document.write( "e)y=-x^2+3x<BR>\n" );
document.write( ".<BR>\n" );
document.write( "By now you should be able to tell that this is a parabola that is a lot like the one in <BR>\n" );
document.write( "problem a) above. It has a minus sign on the x^2 term meaning that the curve rises to<BR>\n" );
document.write( "a peak and then falls back down as you move to the right.  How can you tell the difference<BR>\n" );
document.write( "though between this graph and the one in problem a)?  How about setting y = to zero<BR>\n" );
document.write( "and then solving for the two values of x. If you follow the same process as you did in<BR>\n" );
document.write( "problem a) and problem c) you will find that when y equals zero the values of x that come<BR>\n" );
document.write( "out by factoring that x = 0 is one value and x = +3 is the other.  So the graph has<BR>\n" );
document.write( "(0,0) and (3,0) as two points on the graph.  You now know that the peak occurs where x<BR>\n" );
document.write( "is midway between 0 and 3 on the x-axis.  This midway point is x = 1.5 and you can,<BR>\n" );
document.write( "if you want to, substitute 1.5 in for x in the equation and find the corresponding <BR>\n" );
document.write( "value of y at the peak.<BR>\n" );
document.write( ".<BR>\n" );
document.write( "f)y=x^2+1<BR>\n" );
document.write( ".<BR>\n" );
document.write( "This is a lot like problem a) except this time the x^2 term has positive sign.  This tells<BR>\n" );
document.write( "you that the parabola is \"bowl-shaped\" in that it falls to a low point and then rises back<BR>\n" );
document.write( "upward as you move to the right.  One thing you can do easily is to set x equal to zero<BR>\n" );
document.write( "and you find that the corresponding value of y is +1.  So you know that (0,1) is on the<BR>\n" );
document.write( "graph.<BR>\n" );
document.write( "If you set y = 0 and then try to solve for x you run into a problem because you get<BR>\n" );
document.write( "x^2 = -1.  But you can't solve this by taking the square root of both sides because<BR>\n" );
document.write( "there is no real number that can be multiplied by itself to give -1.  This tells you that<BR>\n" );
document.write( "the graph does not cross or touch the x-axis.<BR>\n" );
document.write( ".<BR>\n" );
document.write( "g)y=x+1<BR>\n" );
document.write( ".<BR>\n" );
document.write( "Has to be a straight-line graph because x terms are only first order (exponent of 1).  <BR>\n" );
document.write( "This is in the slope intercept form.  The slope (multiplier of x) is +1 so the graph<BR>\n" );
document.write( "goes up and to the right (positive slope).  It crosses the y-axis at +1.  You can set<BR>\n" );
document.write( "y = 0 and solve for x to find that the graph crosses the x-axis at x = -1.<BR>\n" );
document.write( ".<BR>\n" );
document.write( "h)y=2x^2<BR>\n" );
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document.write( "You can tell from this equation that when y is set equal to zero, then x must also equal<BR>\n" );
document.write( "zero.  So the point (0,0) is on the graph.  Also when x is set equal to zero, then<BR>\n" );
document.write( "y must also equal zero.  Therefore, (0,0) is the low point on the graph. You can also<BR>\n" );
document.write( "tell that the graph is a parabola (because the x is squared) and that the graph is bowl<BR>\n" );
document.write( "shaped opening upward because the sign on the x-squared term is positive. You could also<BR>\n" );
document.write( "let x = 1 and solve for y to find that y will equal 2 when x equals 1. So (1,2) is a <BR>\n" );
document.write( "point on the graph.  Also by letting x = -1, you get that y equals 2. So (-1,2) is also<BR>\n" );
document.write( "a point on the graph.  If you think about it, this tells you that the y-axis is the<BR>\n" );
document.write( "\"centerline\" of this graph.<BR>\n" );
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document.write( "Hope the above explanations help you to think about the characteristics of graphs of<BR>\n" );
document.write( "parabolic and straight-line equations. And I hope this is what you were looking for in<BR>\n" );
document.write( "terms of how to match the given equations to graphs given in the book. </SPAN>\n" );
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