Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (-3,1) and (3,1) and a y-intercept of -1.

Question

Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (-3,1) and (3,1) and a y-intercept of -1.

y =

Answer 1

Hi Zena K.,

So you must start by observing that the graph is symmetric about the y-axis. That property tells you immediately that the equation MUST be of the form y = ax^4 + bx^3 + cx^2 + dx + e, where b and d are zero! Otherwise those terms would skew the curve right vs. left, and also alter the two maxima's symmetry across the y-axis.

The rest is just a grind in first derivatives; since y' = 4ax^3 + 2cx and the two maxima have slope = y' = 0, then first of all 4a(3^3) = -2c(3) . i.e. 108a = -6c , c = -18a

then you just need to solve the original equation for the two maxima, given that e = -1 (the other x terms drop out at x=0, to give you the y-intercept) : y = 1 = a(3^4) - 18a(3^2) - 1 . I suspect you can take it from there?

Remember that (usually) each of the pieces of information given in a problem has a specific use; your job (given some experience) is to rank those in order of importance in providing answers. Here, the symmetry of the function was the most important piece of information, since until b and d are understood to be zero, you cannot make any headway on solving for any of the other coefficients -- there are simply too many unknowns!

-- Cheers, -- Mr. d.