Find a formula for the fourth degree polynomial p(x) whose graph is symmetric about the y-axis, and which has a y-intercept of 1, and global maxima at (4,257) and (−4,257).
p(x)=

Question

Find a formula for the fourth degree polynomial p(x) whose graph is symmetric about the y-axis, and which has a y-intercept of 1, and global maxima at (4,257) and (−4,257).
p(x)=

anonymous · Answer

@ash2326 @hoblos help please

anonymous · Answer

@AccessDenied @mathmale @texaschic101 @cwrw238 @nincompoop help please!!

kainui · Answer

"fourth degree polynomial"

So we start out with $$p(x)=Ax^4+Bx^3+Cx^2+Dx+E $$
"symmetric about the y-axis"

That means it's an even function, so you have to get rid of all the odd terms, since those aren't symmetric. Play around with them to see what I mean. So you can set those coefficients to 0.

$$p(x)=Ax^5+Cx^2+E $$

"y-intercept of 1"

So what is a y-intercept? This is just the value y has when you're touching it, and that's when x=0. So literally, this is the point (0,1). We can solve for one of our coefficients by plugging in these numbers!

$$1=A0^4=+C0^2+E$$$$1=E$$$$p(x)=Ax^4+Cx^2+1$$

"global maxima at (4,257) and (−4,257)."

Since there is are two global maxima we know that the graph will look something like this: |dw:1395520407338:dw| Since we aren't given any more extra info, we know we can easily make a graph like this without the Cx^2 term so we can just get rid of it, unless you want the entire infinite class of functions that satisfy these conditions. Remember from our picture that it's pointing down, not up, so we add the minus sign to the coefficient on Ax^4 term.

$$p(x)=-Ax^4+1$$

Now plug in one of the conditions.

$$257=A4^4+1$$
You get A=1

$$p(x)=-x^4+1$$
That should be it, if you wanted the infinite number of possibilities or something that looks more like linear algebra I can show you that too.