# Thread: I can NOT figure out what b is!

1. ## 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?

2. 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 $\displaystyle x$, and the intercept of $\displaystyle -2$ means that it is of the form:

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

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

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

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

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

for any $\displaystyle c>0$.

CB