| 
 
 
| Question 65159This question is from textbook algebra for college students
 :  Hello,
 how do i fint the vertex and intercepts for this quadratic function:
 f(x) = x^2 - 6x
 
 thank you! 
This question is from textbook algebra for college students
 
 Found 3 solutions by  josmiceli, funmath, cristiana:
 Answer by josmiceli(19441)
      (Show Source): Answer by funmath(2933)
      (Show Source): 
You can put this solution on YOUR website! f(x) = x^2 - 6x There are a couple of ways to find the vertex, if your teacher is putting the equation in vertex form, let me know and I'll redo this.
 When a quadratic equation is in this form: f(x)=ax^2+bx+c, I prefer using this formula to find the x coordinate of the vertex:
  a=1, b=-6 and c=0
 
  
  
  To find the y-coordinate find f(3)
 
  
  
  The vertex (x,y)=(3,-9)
 You find the y-intercept, by letting x=0:
 
  
  
  The y-intercept is (0,0)  The origin will be both an x and a y-intercept.
 To find the x-intercept, let f(x)=0 and solve for x.
 
  
  x=0 and x-6=0
 x=0 and x=6
 The x-intercepts are (0,0) and (6,0)
 Here's what it looks like:
 
  Happy Calculating!!!
Answer by cristiana(10)
      (Show Source): 
 | 
  
 | 
 |