document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=71862>Question 71862</A>: <I><SPAN STYLE=\"word-break: break-all;\"> My daughter brought this problem home. She seems lost and I have no clue how to help her. My math skills are really rusty it seems. Here's the problem:\r<BR>\n" );
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document.write( "In order to hunt hippopotamuses, a hunter must have a hippopotamus hunting license. Since the hunter can sell the hippos he catches, he can use the proceeds to pay for part or the entire license. If he catches only 3 hippos, he is still in debt by $2050. If he catches 7 hippos he makes a profit of $1550. The African Game and Wildlife Commission allows a limit of 10 hippos per hunter.\r<BR>\n" );
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document.write( "We're asked to define the independent and dependent variables. I figured that the profit is a dependent upon the number caught.<BR>\n" );
document.write( "The cost of the license is independent - right?<BR>\n" );
document.write( "Where I can't even get a clue is how to plot points on a straight line for this problem<BR>\n" );
document.write( "and<BR>\n" );
document.write( "how to find the y intercept and slope of the line.<BR>\n" );
document.write( "please help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #51387 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=51387\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Draw a set of coordinate axes.  On the vertical axis label each major mark in $500 units <BR>\n" );
document.write( "so that from the origin you can go up about 10 units (representing up to the value +$5000).<BR>\n" );
document.write( "Also go down about 10 units (representing down to the value -$5000). On the horizontal<BR>\n" );
document.write( "axis number from the origin in units of 1 ... and go out to 10 or 11 units. <BR>\n" );
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document.write( "Now we are good to go.  The horizontal axis is the independent variable units and these<BR>\n" );
document.write( "units represent the number of hippos ... from 0 to 10 hippos.<BR>\n" );
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document.write( "The vertical axis is the dependent variable dollars, and it is where you plot the  loss or gain <BR>\n" );
document.write( "depending on how many hippos you capture.<BR>\n" );
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document.write( "You are given two points on the graph ... you know that if you catch 3 hippos you are<BR>\n" );
document.write( "still in debt by $2050 so you represent that as minus $2050 0r -$2050. On the x-axis start<BR>\n" );
document.write( "at zero and count out +3 to the right. With your pencil at that point count vertically<BR>\n" );
document.write( "down to the point where you are at -$2050.  At $500 each unit you need to count down 4.1 units.<BR>\n" );
document.write( "Stop. Mark that point.  Call it (3, - 2050) because it represents 3 units along x and <BR>\n" );
document.write( "2050 dollars below the break-even point.<BR>\n" );
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document.write( "Now do the same thing with the other point you are given.  At 7 hippos you will be 7 units<BR>\n" );
document.write( "from zero on the x-axis, and from that point you need to go vertically up to the representation<BR>\n" );
document.write( "for $1550.  Since you are above the x-axis the money is in the positive area and it <BR>\n" );
document.write( "represents a gain.<BR>\n" );
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document.write( "When you have these two points plotted, put a straight-edge along the two points and<BR>\n" );
document.write( "draw a straight line through the two points ... extending this line in both directions<BR>\n" );
document.write( "beyond the two points.<BR>\n" );
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document.write( "Good enough.  Now you have the graph of the line.  The problem asks you to find the slope<BR>\n" );
document.write( "of this line.  Easy enough.  Lightly draw a horizontal line through the point (3, -$2050) <BR>\n" );
document.write( "and then lightly draw a vertical line down from the point (7, +$1550). Notice how these<BR>\n" );
document.write( "two lines intersect to form a right angle.  The slope of the graph is the ratio of the <BR>\n" );
document.write( "lengths of two line segments ... it is the length of the line from the point (7, $1550) down <BR>\n" );
document.write( "to the right angle intersection divided by the length of the line from the point (3, -$2550)<BR>\n" );
document.write( "to the right angle intersection point.  Let's find the length of the line down from the<BR>\n" );
document.write( "point (7, +$1550) vertically down to the intersection point.  At the top end of this line<BR>\n" );
document.write( "the y-value is +$1550 and at the bottom end of this line the corresponding y-value <BR>\n" );
document.write( "is - $2050.  So the total length of the line is $1550 + $2050 = $3600.  [You can also look<BR>\n" );
document.write( "at this as subtracting -$2050 from +$1550 ... and [$1550 - (-$2050)] is also $3600.<BR>\n" );
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document.write( "Next let's find the length of the line from the point (3, -$2050) to the right angle <BR>\n" );
document.write( "intersection.  Can you see that at the left end (at the point 3,-$2050) the horizontal<BR>\n" );
document.write( "value is 3 hippos?  Then at the other end of this line (at the right angle intersection)<BR>\n" );
document.write( "the horizontal value is 7.  So this horizontal length runs from 3 to 7 hippos, and<BR>\n" );
document.write( "therefore the length of the line is 4 ... (7 - 3 = 4).<BR>\n" );
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document.write( "Now you are ready to find the slope by dividing the rise ($3600) by the horizontal <BR>\n" );
document.write( "run (+4) and when you do, the answer will be $900 per hippo.  That means that for<BR>\n" );
document.write( "every hippo captured you will gain $900.  Remember this slope. Note that the<BR>\n" );
document.write( "slope is<BR>\n" );
document.write( "positive if the graph goes up as you move to the right.  If the graph goes down as you<BR>\n" );
document.write( "move to the right the slope is negative.  In this problem, therefore, the slope is positive<BR>\n" );
document.write( "... up and to the right. <BR>\n" );
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document.write( "You can write an equation for the line.  One form of equation for a straight line is called<BR>\n" );
document.write( "the slope intercept form. This form is:<BR>\n" );
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document.write( "y = mx + b<BR>\n" );
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document.write( "Remember that your dollar values are on the y-axis and your hippo values are on the x-axis.<BR>\n" );
document.write( "In this equation form m is the slope and b is the point where the graph crosses the y-axis.<BR>\n" );
document.write( "Let's put the slope into the equation:<BR>\n" );
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document.write( "y = 900x + b<BR>\n" );
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document.write( "We can find the value of b by substituting the values for one of the two given points<BR>\n" );
document.write( "into this equation.  Let's take the point (7, 1550).  So in our equation we are going<BR>\n" );
document.write( "to substitute 7 for x and 1550 for y.  When we do the equation becomes:<BR>\n" );
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document.write( "1550 = (900*7) + b<BR>\n" );
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document.write( "Simplifying this by doing the multiplication results in:<BR>\n" );
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document.write( "1550 = 6300 + b<BR>\n" );
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document.write( "Subtract 6300 from both sides gives you:<BR>\n" );
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document.write( "-4750 = b<BR>\n" );
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document.write( "This tells us that our graph intersects the y-axis at the value -$4750.  Wow.  That has<BR>\n" );
document.write( "to be the cost of the license, because that's how much we spent when we still have<BR>\n" );
document.write( "zero hippos captured.  This should agree with your graph.<BR>\n" );
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document.write( "When we substitute this value for b into the equation we get:<BR>\n" );
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document.write( "y = 900x - 4750<BR>\n" );
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document.write( "as the equation for our line.<BR>\n" );
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document.write( "Notice that where the graph crosses the x-axis the dollar value is zero.  This means that<BR>\n" );
document.write( "you have reached the break-even point and the value on the x-axis is the number of hippos<BR>\n" );
document.write( "that you have to capture to break even. You can solve this exactly by letting y = 0 in<BR>\n" );
document.write( "the equation and solving for x ... as follows:<BR>\n" );
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document.write( "0 = 900x - 4750<BR>\n" );
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document.write( "Adding 4750 to both sides of the equation results in:<BR>\n" );
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document.write( "900x = 4750<BR>\n" );
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document.write( "and dividing both sides by 900 tells you that the graph crosses the x-axis at:<BR>\n" );
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document.write( "x = 4750/900 = 5.27 hippos.<BR>\n" );
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document.write( "Hmmm! That means that when you have captured 5 hippos you still are a little behind.<BR>\n" );
document.write( "But when you have captured 6 hippos you have made a little money. And when you have<BR>\n" );
document.write( "captured 7 hippos you have made $1550.  The problem says that you can only capture<BR>\n" );
document.write( "10 hippos.  And we now know that for each additional hippo we make $900 more.  So what's <BR>\n" );
document.write( "the maximum we can earn due to the limit of ten hippos?  We can count it out.  If we have<BR>\n" );
document.write( "made $1550 for 7 hippos and then we capture the 8th hippo, we add $900 to the $1550 and<BR>\n" );
document.write( "then have $2450.  The 9th hippo gets us another $900 so we're at $3350. And finally,<BR>\n" );
document.write( "the 10th hippo gets us our last $900 for a total of $4250.<BR>\n" );
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document.write( "We also could have used our equation, substituting 10 for x:<BR>\n" );
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document.write( "y = (900)*10 - 4750<BR>\n" );
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document.write( "y = 9000 - 4750 = $4250<BR>\n" );
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document.write( "Hope this helps you decipher what's going on in this problem.  An the above probably<BR>\n" );
document.write( "explains more about hippo hunting than you ever wanted to know.  Follow it through<BR>\n" );
document.write( "carefully and it may help provide some insight into linear graphs and equations.      </SPAN>\n" );
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