A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Just another problem:
A quadratic polynomial \( P(x) \) is such that \( P(x) \) never takes any negative values and \( P(0)=8 \) and the \( P(8)=0\). Find \( P(4) \)
Genre: algebra precalculus
Rating: Easy
anonymous
 4 years ago
Just another problem: A quadratic polynomial \( P(x) \) is such that \( P(x) \) never takes any negative values and \( P(0)=8 \) and the \( P(8)=0\). Find \( P(4) \) Genre: algebra precalculus Rating: Easy

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zarkon: could you please explain with steps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is just too easy for Zarkon :)

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0you did clasify this as easy :)

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.7Let \(P(x)=ax^2+bx+c\), then \(P(0)=8 \implies c=8\) and \(P(8)=0 \implies 8^2a+8b+8=0 \implies b=8a1\). Now, \(P(x)=ax^2(8a+1)x+8\ge 0 \implies (x8)(ax1)\ge 0 \implies a=\frac{1}{8}.\) Thus \(P(4)=\frac{1}{8}(4)^22(4)+8=2+8+8=18.\)

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0i used the fact that (8,0) must be the vertex to get the 2nd equation

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.7Oh that's better I think.

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0Let the quadratic polynomial be \[P(x)=ax^2+bx+c\] as it's given it doesn't take negative values D<0 \[b^24ac<0\] let's substitute x=0 P(0)=8 c=8 P(8)=64a+8b+8=0 or \[8a+b+1=0\] we have b^24ac<0 or b^232a<0 or b^2<32 a we have 8a=1b b^2<44b b^2+4b+4<0 (b+2)^<0 so b is a complex no. of the form =2xi now let x be 1 so \[P(x)=\frac{(1+i)}{8} x^2+(2+i)x+8\] Could anyone point out my mistake??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Seems I am only one who used Maximaminima.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maximaminima : u mean by taking the second derivatives of the eqs and substituting as rts^2 ??

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.7We can also use the derivative (I don't know if that what you mean @Fool). As Zarkon said, the polynomial must have its minimum at x=8. So \(P′(x)=2ax+b=0 \text{ at } x=8.\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes Mr.Math you got that right :)

phi
 4 years ago
Best ResponseYou've already chosen the best response.0@ash if the discriminant is less than zero the square root is imaginary and there are ZERO xintercepts however,the problem states P(8)= 0. so we must have a repeated root at x=8, and the discriminant = 0 Zarkon's approach seems to fastest: given 8 is a root, the equation is y= a(x8)^2 + b 0= a(88)^2 + b > b=0 8= a(8)^2 > a= 1/8 y= (1/8)*(412)^2 = 144/8 = 18
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.