find the values problem of polynomial, please help!

[efdv13]

. Let P(x) = x^4+ ax^3+ bx^2+ cx + d. The graph of y = P(x) is symmetric with respect to the y-axis, hasa relative maximum at (0, 1) and has an absolute minimum at (q, -3).a. Determine the values of a, b, c, and d, and using these values write an expression for P(x).b. Find all possible values of q

all right so i took the deriv, and tried drawing it on my calc, but i honestly have no clue. we've been doing integrals so im really lacking practice in this area, and this is one problem we have as practice for the exam in the spring. Any help in the right direction would be appreciated

 

[efdv13]

so i assume i set the eqaution equal to its negative counter part since its symmetrical f(x)=-f(x)

 

[o_O]

Just skimming through this but play around with it a little. You are given a few clues:

P(x) = P(-x) (Even functions reflected across the y-axis)
P(0) = 1 (It tells you (0,1) is a solution to P(x))
P'(0) = 0

I'm sure you can get something out of that. So play around with it and show your work if you post again with further questions.

 

[efdv13]

all right so i figured out that a and c since they are odd, the only way f(x) equals its opposite is if a and c both equal 0. im left with x^4+bx^2+d, then if i set the equation equals f(0) i get f(0)=d which means d is 1. i then took the derriv and found that x = plus/minus square root of -1/2. im kinda stuck now

 

[Deleted member 4993]

all right so i figured out that a and c since they are odd, the only way f(x) equals its opposite is if a and c both equal 0. im left with x^4+bx^2+d, then if i set the equation equals f(0) i get f(0)=d which means d is 1. i then took the derriv and found that

x = plus/minus square root of -1/2. <<<< what does this mean?

im kinda stuck now

Now the function is:

f(x) = x^4 + bx^2 + 1

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

f'(x) = 0 at x = 0 <<< does not give any new information

Then what??

What other information do we have?

 

[skeeter]

so far, you have P(x) = x[sup:8ektasub]4[/sup:8ektasub] + bx[sup:8ektasub]2[/sup:8ektasub] + 1

you have an absolute max at (q, -3) ...

P(q) = -3 = q[sup:8ektasub]4[/sup:8ektasub] + bq[sup:8ektasub]2[/sup:8ektasub] + 1

P'(q) = 0 = 4q[sup:8ektasub]3[/sup:8ektasub] + 2bq

since q does not equal 0, you can determine that b = -2q[sup:8ektasub]2[/sup:8ektasub] from the second equation.

can you sub for b in the first equation and solve for q ... then find b?