document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=709494>Question 709494</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you do this problem?? I'm so lost! I don't even know where to begin. Thanks in advance..<BR>\n" );
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document.write( "The viscosity (resistance to flow) of normal motor oil decreases as its temperature increases. \"All weather\" motor oils retain roughly the same viscosity over a range of temps. Suppose that the viscosity of low-40 is given by M= - 4T^3 + 15T^2 - 12T + 185 for 0 &#8804; T &#8804; 3.<BR>\n" );
document.write( " M=viscosity in centipoise (cP)<BR>\n" );
document.write( " T= Temp in hundreds of degrees\r<BR>\n" );
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document.write( " A. Find the minimum and maximum viscosity in the domain and the temperature at which they occur.\r<BR>\n" );
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document.write( " B. Show that the function has 1 real zero, but not in the given domain.\r<BR>\n" );
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document.write( " C. By synthetic substitution, factor the cubic expression into a linear times a quadratic. Show that the quad has no real zeros. How does this fact agree with the graph of this problem? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #436629 by </B><A HREF=/tutors/aboutme.mpl?userid=josh_jordan><B>josh_jordan(263)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josh_jordan><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/cgi-bin/show-face.mpl?user=josh_jordan><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=436629\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's look at the formula given in this word problem, and determine what it's telling us. It tells us the following:\r<BR>\n" );
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document.write( "Viscosity = -4(temperature)^3 + 15(temperature)^2 - 12(temperature) + 5, and that the temperature must be greater than or equal to 0, and less than or equal to 3, where the Viscosity is labeled as M, and the temperature is labeled as T, in hundreds of degrees\r<BR>\n" );
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document.write( "Part A wants us to find the minimum and maximum viscosity in the domain, and the temperature in which they occur.  In other words, they want us to plug in the numbers 0 and 3 into the equation for T, to find the viscosity. 0 is the minimum temperature we can plug in, and 3 is the maximum temperature we can plug in.  So, let's plug in the number 0 into our equation:\r<BR>\n" );
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document.write( "M = -4(0)^3 + 15(0)^2 - 12(0) + 185 -----> M = 0 + 0 - 0 + 185.  Therefore, when the temperature is 0 degrees, the viscosity is 185cP\r<BR>\n" );
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document.write( "Now, let's plug in 3:\r<BR>\n" );
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document.write( "M = -4(3)^3 + 15(3)^2 - 12(3) + 185 -----> M = -4(27) + 15(9) - 36 + 185 -----> M = -108 + 135 - 36 + 185 -----> M = 176cP\r<BR>\n" );
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document.write( "So, the answer to A is 0 degrees = 185cP and 300 degrees = 176cP\r<BR>\n" );
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document.write( "B wants us to show that our function has one real zero, but not in the domain, meaning our one real zero is not equal to 0, 1, 2, or 3.  To find our one real zero, we need to synthetically divide our equation by our possible zeros. To find our possible zeros, we need to perform the rational zero test, which is where we will divide all of the factors of our constant, in this case, 185, by all of the factors of our leading coefficient, in this case, 4:\r<BR>\n" );
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document.write( "(+-1, +-5, +-37, +-185)/(+-1, +-2, +-4), which gives us our possible zeros we need to test:  +-1, +-1/2, +-1/4, +-5, +-5/2, +-5/4, +-37, +-37/2, +-37/4, +-185, +-185/2, +-185/4.  Remember that our one real zero cannot be 0,1,2, or 3, so we will not have to test out +-1.  To save some time, and typing, I will tell you that 5 is the real zero.  Once you synthetically divide our equation for M by 5, you will see that 5 is the real zero.\r<BR>\n" );
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document.write( "So, the answer to B is discovered by using the rational zero test to discover the possible zeros, then eliminating 1 from that possibility, since 1 is part of the given domain, and finally, synthetically dividing the given polynomial by 5.\r<BR>\n" );
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document.write( "C wants you to factor this equation into a linear times a quadratic using synthetic substitution, then show that the quadratic has no real zeros, then explaining how a graph shows this fact.\r<BR>\n" );
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document.write( "To use synthetic substitution to factor, we will take the real zero we found from B, place it on the left, and write all of the coefficients of the original formula, to the right side of the real zero, separated by a line.  In other words, it should look like this (you are using the same process as synthetic division):\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \text {5    |   -4   15   -12   185}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \text {             -20   -25  -185}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \text {--------------------}\"><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \text {         -4  -5   -37   0}\">\r<BR>\n" );
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document.write( "Now that we have used synthetic substitution, we can convert our zero, and the quotient, into factor form:\r<BR>\n" );
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document.write( "(x - 5)(-4x^2 - 5x - 37)\r<BR>\n" );
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document.write( "We are now asked to show that the quadratic factor has no real zeros.  All you need to do here is either use the rational zero test to determine the possible zeros, then synthetically divide the quadratic by each zero OR you can use the quadratic formula to show that there are no real zeros as solutions.\r<BR>\n" );
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document.write( "Finally, you are asked how this fact agrees with graph of the problem.  When you draw a graph of the original formula, you will see that the function crosses the y axis once, at point (0,185) and the x axis once, at point (5,0).  Since there are no more instances where the function crosses the x or y axis, there cannot be any more real zeroes, in neither the quadratic factor OR the original formula.\r<BR>\n" );
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document.write( "Hope this helps and that I didn't confuse you more!<BR>\n" );
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