document.write( "Question 495651: Explain why the graph of a polynomial function with real coefficients must have a y-intercept but may have no x-intercept. \n" ); document.write( "

Algebra.Com's Answer #336139 by oberobic(2304) $\"\"$ $\"About$

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The graph could be shift up or down such that it never touches x.
\n" ); document.write( "Consider the following graphs:
\n" ); document.write( "y = x^2 +1.
\n" ); document.write( " $\"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C+x%2Ax%2B1%29\"$
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\n" ); document.write( "y = -x^2 -1.
\n" ); document.write( " $\"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C+-x%2Ax-1%29\"$
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\n" ); document.write( "The \"why\" part of your question is because the function is define for all 'x', from -infinity to +infinity.
\n" ); document.write( "So, by definition, it is defined for x=0. That is the y-intercept.
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\n" ); document.write( "As I showed above, you can manipulate the values to avoid the x-axis, at least for even polynomials. But with odd polynomials, including linear equations (the exponent = 1), there will be an x-intercept.
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\n" ); document.write( "Consider:
\n" ); document.write( "y = x^3 + 1
\n" ); document.write( " $\"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C+x%5E3%2B1%29\"$
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