document.write( "Question 4114: Find the vertex and intercepts for each quadratic function. Does it open up or down? y = x^2 – 4
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Algebra.Com's Answer #1824 by rapaljer(4671) $\"\"$ $\"About$

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There are several ways to find the vertex and graph a quadratic function. One easy way is to realize that in the general case\r
\n" ); document.write( "\n" ); document.write( " $\"+y+=+ax%5E2+%2B+bx+%2B+c+\"$ \r
\n" ); document.write( "\n" ); document.write( "the vertex is at \r
\n" ); document.write( "\n" ); document.write( " $\"+x=+-b%2F2a+\"$ \r
\n" ); document.write( "\n" ); document.write( "Therefore for $\"y+=+x%5E2+-+4\"$ the coefficient of x is zero, so the vertex is at x=0. When x = 0, y = -4, so the vertex is at (0, -4).\r
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\n" ); document.write( "\n" ); document.write( "The graph opens upward, since the coeffient of x^2 is positive. It would open downward if the coefficient of x^2 is negative.\r
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\n" ); document.write( "\n" ); document.write( "The y-intercept is where x=0, which is at the point (0, -4).\r
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\n" ); document.write( "\n" ); document.write( "The x-interecepts are where y=0, which are at \r
\n" ); document.write( "\n" ); document.write( " $\"0=x%5E2+-+4\"$ or x=2 and at x=-2.\r
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