document.write( "Question 203135: 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 worky
\n" ); document.write( "y = -x^2 + 2x - 1 \n" ); document.write( "
Algebra.Com's Answer #153256 by jsmallt9(3758)\"\" \"About 
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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+%2B+2x+-+1\"
\n" ); document.write( "Since solving quadratic equations by factoring is easier than using the quadratic formula and since factoring is easier when the leading (first) coefficient is 1, I am going to start by factoring out a -1:
\n" ); document.write( "\"0+=+-1%28x%5E2+-+2x+%2B+1%29\"
\n" ); document.write( "This factors easily:
\n" ); document.write( "\"0+=+-1%28x+-+1%29%5E2\"
\n" ); document.write( "The only way for a product to be zero is if one of the factors is zero. The only factor that could be zero is (x-1). So the only solution to \"0+=+-1%28x+-+1%29%5E2\" is the solution to (x - 1) = 0, which is x = 1. So there is only one x-intercept: (1, 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, with just one x-intercept, is in the last category. Its vertex is not above or below the x-axis. It is on the x-axis. It is the x-intercept: (1, 0). \n" ); document.write( "
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