November 3rd, 2019, 06:23 PM  #1 
Newbie Joined: Nov 2019 From: California Posts: 5 Thanks: 0  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

November 3rd, 2019, 06:36 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 
symmetric over the yaxis $\implies$ f(x) is an even quartic function ... $f(x) = ax^4 + bx^2 + c$ yintercept = 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 $b$ 
November 3rd, 2019, 06:51 PM  #3 
Newbie Joined: Nov 2019 From: California Posts: 5 Thanks: 0 
How exactly can I solve for a and b... What 2 equations do I have to do a system on?

November 3rd, 2019, 07:04 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1473  
November 3rd, 2019, 07:24 PM  #5 
Newbie Joined: Nov 2019 From: California Posts: 5 Thanks: 0 
right? then how can I find a and b to find the function??

November 3rd, 2019, 07:44 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 
$f(x) = ax^4 + bx^2  2$ $f(2) = 16a + 4b  2 = 0 \implies 8a + 2b = 1$ $f'(x) = 4ax^3 +2bx$ $f'(2) = 32a + 4b = 0 \implies 8a + b = 0$ $b = 1$ and $a = \dfrac{1}{8}$ 

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