document.write( "Question 1200631: Function h is defined by h : x —> x^2 + 4x for x ≥ k and it is given that h has an inverse.
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Algebra.Com's Answer #834844 by Edwin McCravy(20062) $\"\"$ $\"About$
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document.write( "Even though the graph of h : x —> x^2 + 4x is this parabola:\r\n" );
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document.write( "since it has a inverse, it must pass the horizontal line test.  But as you see\r\n" );
document.write( "in the graph below, the green horizontal lines cuts the graph in two places, so\r\n" );
document.write( "the function as graphed does not have an inverse. It is not one-to-one.\r\n" );
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document.write( "This is where  comes in. We must chop off enough of the graph, so\r\n" );
document.write( "that what's left will pass the horizontal line test. We will chop it off at\r\n" );
document.write( "the smallest value of x so that no points on the graph will be directly to the\r\n" );
document.write( "left of any other part of the graph.  So obviously we chop it off at the very\r\n" );
document.write( "bottom point, which is the vertex of the parabola. \r\n" );
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document.write( "To find the vertex of a parabola we use the formula for the x-coordinate of the\r\n" );
document.write( "vertex, which is . In the equation  \"a\" is the coefficient of \r\n" );
document.write( "x², which is 1. The \"b\" is the coefficient of x, which is 4.\r\n" );
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document.write( ". Therefore k=-2.\r\n" );
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document.write( "So the x-coordinate of the vertex is -2, when we substitute -2 for x in\r\n" );
document.write( " we get the y-coordinate of the vertex . \r\n" );
document.write( "So the vertex is the point (-2,-4). So we chop the parabola where x=k=2, \r\n" );
document.write( "right at the vertex, \r\n" );
document.write( "so we will only leave the right half of the parabola, like this:\r\n" );
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document.write( ". We only use the part at or right of x=-2\r\n" );
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document.write( "To find the inverse of h(x) = x² + 4x  for x > -2\r\n" );
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document.write( "1. We replace h(x) by y,\r\n" );
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document.write( "2. We replace x by y and replace y by x:\r\n" );
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document.write( "We solve for y:\r\n" );
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document.write( "Multiply through by -1 to make squared term positive:\r\n" );
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document.write( "Use the quadratic formula:\r\n" );
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document.write( "Now we work out the domain of the inverse \r\n" );
document.write( "Replace y by what y equals:\r\n" );
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document.write( "Add 2 to both sides\r\n" );
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document.write( "Since it is non-negative we use the positive sign\r\n" );
document.write( "for the square root. \r\n" );
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document.write( "That tells us that for the inverse we also use the \r\n" );
document.write( "positive sign for the same square root: \r\n" );
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document.write( "Going back to the inequality, \r\n" );
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document.write( "we square both sides. \r\n" );
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document.write( "That's the domain of the inverse function. So the inverse is\r\n" );
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document.write( "although we write h^-1(x) for y:\r\n" );
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document.write( "  <---ANSWER\r\n" );
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document.write( "If you want to use the same kind of notation your teacher uses,\r\n" );
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document.write( "  <---ORIGINAL FUNCTION with k = -2\r\n" );
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document.write( "  <---INVERSE FUNCTION\r\n" );
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document.write( "Here is the graph of the inverse on the same set of axes (in blue):\r\n" );
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document.write( "And you see that the inverse is the reflection of the original function\r\n" );
document.write( "across the identity line, whose equation is y = x  (where x and y are\r\n" );
document.write( "identically equal and the identity line is the line that bisects the 1st and 3rd\r\n" );
document.write( "quadrants (in green, dashed since it's not part of either graph).\r\n" );
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document.write( "Edwin
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