document.write( "Question 143395: Find the equation of the parabola (second degree polynomial) that passes through the points (-3,0), (4,0), and (0,2) \n" ); document.write( "

Algebra.Com's Answer #104357 by solver91311(24713) $\"\"$ $\"About$

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First thing to notice is that two of the points are on the x-axis and that tells us that we have a parabola whose axis is perpendicular to the x-axis, i.e. vertical, and that the x-coordinates of these two points are the zeros of the desired polynomial function.\r
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\n" ); document.write( "\n" ); document.write( "Knowing that we have two zeros for a 2nd degree polynomial, we can derive A quadratic function simply by multiplying $\"%28x%2B3%29%28x-4%29=x%5E2-x-12\"$ . However, this misses the mark because if you calculate the y-intercept you get (0,-12) instead of the desired (0,2).\r
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\n" ); document.write( "\n" ); document.write( "Fortunately, you can multiply any polynomial by any constant and not change the zeros. $\"x%5E2%2B4x-5\"$ , $\"2x%5E2%2B8x-10\"$ , and $\"x%5E2%2F4%2Bx-5%2F4\"$ all have the same zeros, namely 1 and -5.\r
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\n" ); document.write( "\n" ); document.write( "So for the function in question, we need to answer, \"What can we multiply by so that the constant term will be 2?\" Answer: $\"-1%2F6\"$ .\r
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\n" ); document.write( "\n" ); document.write( "So: $\"f%28x%29=-x%5E2%2F6%2Bx%2F6%2B2\"$ is the desired function.\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( " $\"f%280%29=-%280%29%5E2%2F6%2B0%2F6%2B2=2\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"f%28-3%29=-%28-3%29%5E2%2F6%2B%28-3%29%2F6%2B2=-9%2F6-3%2F6%2B2=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"f%284%29=-%284%29%5E2%2F6%2B%284%29%2F6%2B2=-16%2F6%2B4%2F6%2B2=0\"$ \n" ); document.write( "

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