# Thread: Problem on finding values

1. ## Problem on finding values

I'm having trouble with doing this problem, and I'm not really sure how to even first go about it. All I was able to figure out is that $d=1$. If someone could explain this problem to me step by step that would be really appreciated, thank you!

Let $P(x)=x^4+ax^3+bx^2+cx+d$. The graph $y=P(x)$ is symmetric with respect to the Y-axis, has relative maximum at $(0,1)$, and has an absolute minimum at $(q, -3)$

A) Determine the values $a, b, c, d$ and using these values write an expression for $P(x)$

B) Find all possible values of $q$

2. Originally Posted by choi_siwon
I'm having trouble with doing this problem, and I'm not really sure how to even first go about it. All I was able to figure out is that $d=1$. If someone could explain this problem to me step by step that would be really appreciated, thank you!

Let $P(x)=x^4+ax^3+bx^2+cx+d$. The graph $y=P(x)$ is symmetric with respect to the Y-axis, has relative maximum at $(0,1)$, and has an absolute minimum at $(q, -3)$

A) Determine the values $a, b, c, d$ and using these values write an expression for $P(x)$

B) Find all possible values of $q$
Since it has a relative max at (0,1)

$P'(0)=0$

$P'(x)=4x^3+3ax^2+2bx+c \implies P'(0)=0=c$

If P is symmetric with the y axis then

$P(x)=P(-x)$ for ALL values of x.

$x^4+ax^3+bx^2+1=x^4-ax^3+bx^2+1$

$2ax^3=0$ this can only be zero for all values of x if a=0

$P(x)=x^4+bx^2+1 \implies P'(x)=4x^3+2xb$

$q=P(-3)=81+9b+1$

$P'(-3)=0=-108-6b \implies b=-\frac{54}{3}=-18$

$q=81-162+1=-80$

3. Thanks for the link skeeter!

And thank you so much TheEmptySet! I understand it now.