document.write( "Question 203134: Without drawing the graph determine how many x- intercepts the parabola has and whether its vertex lies above or below or on the x axis Show your work
\n" ); document.write( "y = x^2 - 5x + 6 \n" ); document.write( "
Algebra.Com's Answer #153257 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
X-intercepts are where a a graph intersects the x-axis. In other words, x-intercepts are x-values which make the function value 0. So to find x-intercepts we need to find the x-value(s) where f(x) = 0.

\n" ); document.write( "So finding the x-intercepts of your function means solving:
\n" ); document.write( "\"0+=+x%5E2+-+5x+%2B+6\"
\n" ); document.write( "Factoring this we get:
\n" ); document.write( "\"0+=+%28x+-+2%29%28x+-+3%29\"
\n" ); document.write( "The only way for a product to be zero is if one of the factors is zero. So the solution to \"0+=+%28x+-+2%29%28x+-+3%29\" is: \"%28x+-2%29+=+0\" or \"%28x+-+3%29+=+0\". The solutions to these are: x = 2 or x = 3.
\n" ); document.write( "There are two x-intercepts: (2, 0) and (3, 0)

\n" ); document.write( "As for where the vertex is located, think about parabolas in general. In general the possibilities are:
  • The parabola is completely above the x-axis. These parabolas will have no x-intercepts and the leading coefficient is positive.
  • The parabola is completely below the x-axis. These parabolas will have no x-intercepts and the leading coefficient is negative.
  • Part of the parabola is above the x-axis and part of it is below the x-axis. These parabolas have 2 x-intercepts. If the leading coefficient is positive the vertex will be below the x-axis. If the leading coefficient is negative, the vertex will be above the x-axis.
  • Parabolas whose vertex is not above or below but on the x-axis. These parabolas will have just one x-intercept, the vertex.

\n" ); document.write( "Our parabola has 2 x-intercepts and its leading coefficient is 1 (which is positive). So our vertex is below the x-axis. \n" ); document.write( "
\n" );