document.write( "Question 425396: Find the vertex and intercepts for the quadratic function and sketch the graph.\r
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document.write( "y=x^2+2x-24\r
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document.write( "My thoughts:\r
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document.write( "0=x^2+2x-24....but then I'm not sure where I go from here. And I know once I find x, I need to find y? \n" );
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Algebra.Com's Answer #296232 by scott8148(6628)![]() ![]() You can put this solution on YOUR website! the general vertex form of a quadratic is ___ y = a(x - h)^2 + k ___ (h,k) is the vertex and the \"a\" is related to the shape\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "separating out a perfect square ___ y = x^2 + 2x + 1 - 25 = (x + 1)^2 - 25\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "so the vertex is (-1,-25)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the x-intercepts (roots) are when y equals zero ___ factoring ___ 0 = (x + 6)(x - 4)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the y intercept is when x equals zero ___ y = (0)^2 + 2(0) - 24\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "this gives you four points to sketch the parabola \n" ); document.write( "remember that the vertex is the \"nose\" of the parabola and it is on the axis of symmetry \n" ); document.write( " |