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 $\displaystyle d=1$. If someone could explain this problem to me step by step that would be really appreciated, thank you!

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

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

B) Find all possible values of $\displaystyle q$

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 $\displaystyle d=1$. If someone could explain this problem to me step by step that would be really appreciated, thank you!

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

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

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

$\displaystyle P'(0)=0$

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

If P is symmetric with the y axis then

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

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

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

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

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

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

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

3. Thanks for the link skeeter!

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