document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=227719>Question 227719</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1)If the graph of y=x2-6x+c is tangent to the x-axis, then c equals<BR>\n" );
document.write( "(1) 3<BR>\n" );
document.write( "(2) -3<BR>\n" );
document.write( "(3) 9<BR>\n" );
document.write( "(4) -9\r<BR>\n" );
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document.write( "6)Which parabola intersects the x-axis in two distinct points?<BR>\n" );
document.write( "(1) y=(x+5)^2<BR>\n" );
document.write( "(2) y=(x-5)^2<BR>\n" );
document.write( "(3) y=x^2-25<BR>\n" );
document.write( "(4) y=x^2+25\r<BR>\n" );
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document.write( "9)If the graph of y=4x2+bx=1 intersects the x-axis in two distinct points,then b may be equal to<BR>\n" );
document.write( "(1) 3<BR>\n" );
document.write( "(2) 4<BR>\n" );
document.write( "(3) -4<BR>\n" );
document.write( "(4) 5\r<BR>\n" );
document.write( "<BR>\n" );
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document.write( "These are the problems i cant solve. Please show work and i dont know if this is the right topic but i know it Algebra 2 thanks.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #169187 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=169187\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1)If the graph of y=x2-6x+c is tangent to the x-axis, then c equals...<BR>\n" );
document.write( "The only way for a parabola to be tangent to the x-axis is if the vertex is on the x-axis. And the x-coordinate of the vertex of a parabola is -b/2a, which in this case is -(-6)/2(1) = 3. And on the x-axis the y-coordinate is 0. So all we have to do is substitute 3 for x and 0 for y and solve for c:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%280%29+=+%283%29%5E2+-6%283%29+%2Bc ALIGN=MIDDLE  ALT=\"%280%29+=+%283%29%5E2+-6%283%29+%2Bc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+=+9+-+18+%2B+c ALIGN=MIDDLE  ALT=\"0+=+9+-+18+%2B+c\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+=+-9+%2B+c ALIGN=MIDDLE  ALT=\"0+=+-9+%2B+c\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=9+=+c ALIGN=MIDDLE  ALT=\"9+=+c\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "6)Which parabola intersects the x-axis in two distinct points?<BR>\n" );
document.write( "(1) y=(x+5)^2<BR>\n" );
document.write( "(2) y=(x-5)^2<BR>\n" );
document.write( "(3) y=x^2-25<BR>\n" );
document.write( "(4) y=x^2+25<BR>\n" );
document.write( "Unlike #1, we do not want the vertex on the x-axis. Since all of these equations have an \"a\" (coefficient of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2 ALIGN=MIDDLE  ALT=\"x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >) which is positive, these are all parabolas which open upward. So in order to intersect the x-axis twice, we need a vertex which is below the x-axis. So find the vertex of each. (1) and (2), you will find, have a vertex on the x-axis (like problem #1). And (4) has a vertex above the x-axis. This parabola does not intersect the x-axis at all. Only (3) have a vertex below the x-axis.<br><BR>\n" );
document.write( "9)If the graph of y=4x2+bx=1 intersects the x-axis in two distinct points,then b may be equal to<BR>\n" );
document.write( "(1) 3<BR>\n" );
document.write( "(2) 4<BR>\n" );
document.write( "(3) -4<BR>\n" );
document.write( "(4) 5<BR>\n" );
document.write( "Like problem #6 we need a vertex below the x-axis. And in order for the vertex to be below the x-axis, the y-coordinate of the vertex has to be negative. So what we can do is try each possible b. Then find the x-coordinate of the vertex, -b/2a and then see if the y-coordinate for that x in negative. If we try (1), b=3:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+4x%5E2%2B3x%2B1 ALIGN=MIDDLE  ALT=\"y+=+4x%5E2%2B3x%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "x-coordinate of the vertex: -3/2(4) = -3/8<BR>\n" );
document.write( "y-coordinate of the vertex = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4%28%28-3%29%2F8%29x%5E2%2B3%28%28-3%29%2F8%29%2B1+=+4%289%2F64%29+%2B+%28-9%29%2F8+%2B+1+=+9%2F16+%2B+%28-9%29%2F8+%2B+1+=+9%2F16+%2B+%28-18%29%2F16+%2B+16%2F16+=+7%2F16 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Since the y is positive, this is not our solution.<br><BR>\n" );
document.write( "Repeating this process for the other possible b's we should find that only (4), b=5, results in a y-coodinate which is negative. So only with b=5 do we get a vertex below the x-axis and, therefore, a parabola that intersects the x-axis twice.<BR>\n" );
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