f(x)=1/4(x+2)^2+4
Vertex is___(4,-2)__
The line of symmetry is_x=-2____
What is the max/min value of f(x)_(4,-2)____

Is this correct?

Question

f(x)=1/4(x+2)^2+4
Vertex is___(4,-2)__
The line of symmetry is_x=-2____
What is the max/min value of f(x)_(4,-2)____

Is this correct?

anonymous · Answer

no
vertex = (-2,4)
axis of symmetry: y =-2
min: (-2,4)

anonymous · Answer

you just switched things is all; no worries

anonymous · Answer

;-/ I will try another one...hold on just a sec

anonymous · Answer

Okay

anonymous · Answer

$$y-4=4 \times \frac 4 {16}(x+2)^2 $$

Vertex is at $(-2,4)$
Axis of symmetry  is $ x=-2 $ 

Minimum value is at it's vertex $(-2,4)$; There will be no maximum value as this parabola is concave upward.

anonymous · Answer

No, if it was x = -2, the axis would go through the whole x axis line. the whole point is to divide the parabola

anonymous · Answer

Thus, : y = (-2,4)

anonymous · Answer

The axis of symmetry is parallel to the y-axis. $x=-2 $ is correct.

anonymous · Answer

no:  I think the line proves otherwise...

anonymous · Answer

f(x)=3x^2-12x+15
Vertex: (2,3)
What is the equation of the line of symmetry:  2
What is the max/min:(2,3)

anonymous · Answer

Either you have a problem with understanding axis of symmetry or you don't know how the line $ x=-2 $ and $y=-2$ will look.

anonymous · Answer

@alisa845: Please post it as a new question.-Thanks

anonymous · Answer

Ok