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I'v quistion on Quadratic Functions

  • Thread starter megatronic
  • Start date
  • #1
16
0

Homework Statement


Let f(x)= 125x-6x2 find the maximum value of f to four decimal places graphically


Homework Equations


(x)= 125x-6x2


The Attempt at a Solution


Homework Statement



attempt solution:

x=-b+[tex]\sqrt{}b2-4ac[/tex]/2a

x=-125+[tex]\sqrt{}1252-4(-6)(0)[/tex]/2a

x= 0/-12=0

x=-b-[tex]\sqrt{}b2-4ac[/tex]/2a

x=-b-[tex]\sqrt{}b2-4ac[/tex]/2a

x=(-125-125)/-12

x=-125/6


Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
214
1
Do you know how to make graph for quadratic equations?
 
  • #3
214
1
and also show your attempt otherwise nobody will help you here because of the rules of PF
 
  • #4
16
0
Do you know how to make graph for quadratic equations?
ya I know to make graph for quadratic
 
  • #5
Mentallic
Homework Helper
3,798
94
Ok then, do you realize there is a symmetry of any parabola on either end of its max/min? So if you find the roots of the quadratic, what can you deduce about the coordinates of the max/min from that?
 
  • #6
33,503
5,191

Homework Statement


Let f(x)= 125x-6x2 find the maximum value of f to four decimal places graphically


Homework Equations


(x)= 125x-6x2


The Attempt at a Solution


Homework Statement



attempt solution:

x=-b+[tex]\sqrt{}b2-4ac[/tex]/2a

x=-125+[tex]\sqrt{}1252-4(-6)(0)[/tex]/2a

x= 0/-12=0

x=-b-[tex]\sqrt{}b2-4ac[/tex]/2a

x=-b-[tex]\sqrt{}b2-4ac[/tex]/2a

x=(-125-125)/-12

x=-125/6
What you've done is attempt to find the x-intercepts, which is not at all what this problem is asking for. In addition, there is one intercept that you did not find, and the one you found is incorrect.
 
  • #7
Think what 'maximum' is. Note that maximum does not refer to an x value.

There is a formula to determine the maximum or minimum value of a parabola. Give it a hard think. It'll come to you.
 
  • #8
33,503
5,191
Think what 'maximum' is. Note that maximum does not refer to an x value.

There is a formula to determine the maximum or minimum value of a parabola. Give it a hard think. It'll come to you.
The OP is not supposed to use a formula. He/she is supposed to find the maximum by looking at a graph of the function.
 

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