document.write( "Question 1208802: Find the value of $a$ for which there is exactly one real value of $x$ such that $f(x) = a,$ where
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Algebra.Com's Answer #847254 by ikleyn(52780) $\"\"$ $\"About$

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document.write( "In this problem you have the given parabola and horizontal line y= a.\r\n" );
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document.write( "Without long conversation and discussions, they want you find the vertex of the parabola f(x) = x^2 + 4x - 31.\r\n" );
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document.write( "To find the vertex, complete the square\r\n" );
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document.write( "    x^2 + 4x - 31 = (x^2 + 4x) - 31 = (x^2 + 4x + 4) - 4 - 31 = (x+2)^2 - 35.\r\n" );
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document.write( "This is the vertex form of the given parabola equation, and it shows you that\r\n" );
document.write( "the minimum of the parabola is -35 at the point (x,y) = (-2,-35).\r\n" );
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document.write( "The entire parabola is above the level  y = -35,  having only one point - the vertex,- with this horizontal line y= -35.\r\n" );
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document.write( "The parabola and horizontal line y = -35 have only one common point at the vertex; this point is the tangent point.\r\n" );
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document.write( "for all  b > a,  equation  f(x) = b  has two solutions;\r\n" );
document.write( "for all  c < a,  equation  f(x) = c  has no solutions.\r\n" );
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document.write( "So, a= -35 is the ANSWER\r\n" );
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