SOLUTION: Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the function. f(x)=1/4(x+2)^2+1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the function. f(x)=1/4(x+2)^2+1      Log On


   

Question 411041: Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the function.
f(x)=1/4(x+2)^2+1
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

f(x)=1/4(x+2)^2+1

Using the vertex form of a parabola, y=a%28x-h%29%5E2+%2Bk where(h,k) is the vertex

the vertex: Pt(-2,1)

the line of symmetry: x = -2 

and the maximum or minimum value of f(x): 

f(-2) = 1 is a minimun  |parabola opens upward as a = (1/4)>0





f(x)= -22x^2+2x+6 OR f(x) = -22(x- 1/11)^2 + 64/11 |completing the square