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November 3rd, 2019, 06:23 PM   #1
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Help solving this??

Find the equation of a quartic polynomial whose graph is symmetric about the y -axis and has local maxima at (−2,0) and (2,0) and a y -intercept of -2
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November 3rd, 2019, 06:36 PM   #2
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symmetric over the y-axis $\implies$ f(x) is an even quartic function ...

$f(x) = ax^4 + bx^2 + c$

y-intercept = -2 $\implies c = -2$

$f(x) = ax^4 + bx^2 - 2$

the local extrema info tell you two things...

$f(x) = 0$ at $x=\pm 2$

$f'(x) = 4ax^3 + 2bx = 0$ at $x=\pm 2$

you can now determine $a$ and $b$
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November 3rd, 2019, 06:51 PM   #3
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How exactly can I solve for a and b... What 2 equations do I have to do a system on?
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November 3rd, 2019, 07:04 PM   #4
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Quote:
Originally Posted by takelight View Post
How exactly can I solve for a and b... What 2 equations do I have to do a system on?
I think you'll find you can only solve for one in terms of the other.
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November 3rd, 2019, 07:24 PM   #5
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right? then how can I find a and b to find the function??
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November 3rd, 2019, 07:44 PM   #6
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$f(x) = ax^4 + bx^2 - 2$

$f(2) = 16a + 4b - 2 = 0 \implies 8a + 2b = 1$

$f'(x) = 4ax^3 +2bx$

$f'(2) = 32a + 4b = 0 \implies 8a + b = 0$

$b = 1$ and $a = -\dfrac{1}{8}$
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