# I can NOT figure out what b is!

• Nov 4th 2008, 03:43 PM
purplegirl1818
I can NOT figure out what b is!
Sorry, i posted the wrong thread title... this is my question:

Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (-1, 1) and (1, 1) and a y-intercept of -2.

I found y=-2x^(4)+5x^(2)-2 but that is not correct... any suggestions?
• Nov 5th 2008, 05:11 AM
CaptainBlack
Quote:

Originally Posted by purplegirl1818
Sorry, i posted the wrong thread title... this is my question:

Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (-1, 1) and (1, 1) and a y-intercept of -2.

I found y=-2x^(4)+5x^(2)-2 but that is not correct... any suggestions?

As it is symmetric it has only even powers of $x$, and the intercept of $-2$ means that it is of the form:

$y=ax^4+bx^2-2$

You know that $x= \pm 1$ give maxima so $y'(1)=y'(-1)=0$.

$y'=4ax^3+2bx=0$

so the critical points are $x=0$, $x=\pm \sqrt{b/(2a)}=\pm 1$, so we must have b=2a, and as the $x=\pm 1$ critical points must be maxima we have:

$y=-cx^4-2cx^2-2$

for any $c>0$.

CB