document.write( "Question 147536This question is from textbook algebra
\n" ); document.write( ": It asks, without drawing the graph of the given equation, determine: How many x-intercepts the parabola has,wether its vertex lies above or below or on the x-axis.
\n" ); document.write( "The problem is y=-x^2+2x-1. I have a hard time understanding how to solve it, please help me. Thank you=) \n" ); document.write( "

Algebra.Com's Answer #107924 by mangopeeler07(462) $\"\"$ $\"About$

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Here's an easy solution, step by step, no formulas involved: the x-intercept is whatever x is when y is zero. factor the original equation by taking out negative one and then factoring $\"-%28x-1%29%28x-1%29\"$ . Set it equal to zero. Then solve for each expression individually to equal zero. What minus 1 equals zero? You should have gotten 1 for both. That means the x-intercept is 1. There is only one x-intercept because you got the same solution for x in both expressions. Since there is only 1, it lies on the x-axis.\r
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\n" ); document.write( "\n" ); document.write( "**Just in case you're wondering whether the parabola opens up or down, it opens down. Whenever $\"x%5E2\"$ is positive, it moves up, and vice versa. Think of it this way: If your $\"x%5E2\"$ is positive, then y wants to be positive too, so it moves upward. If your $\"x%5E2\"$ is negative, then y wants to be negative too, so it moves downward.** \n" ); document.write( "

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