SOLUTION: Identify the vertex, axis of symmetry, and direction of opening for: y=1/2(x-8)^2+2. Thank you

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Identify the vertex, axis of symmetry, and direction of opening for: y=1/2(x-8)^2+2. Thank you      Log On


   

Question 41896: Identify the vertex, axis of symmetry, and direction of opening for:
y=1/2(x-8)^2+2.
Thank you
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
First re-arrange the given equation to compare with standard form.
y=%281%2F2%29%28x-8%29%5E2%2B2
or %28y-2%29=4%2A%281%2F8%29%2A%28x-8%29%5E2

Let us transform the equation to the new co-ordinate system (X,Y) such that X=x-8 and Y=y-2.
Then the given equation becomes X%5E2+=+4%281%2F8%29Y.
This is a parabola with the Y-axis as its axis.
Comparing with the standard equation X%5E2+=+4aY, a=1%2F8.

AXIS OF SYMMETRY AND DIRECTION OF OPENING
Here the axis of symmetry is x = 8 and as a+=+1%2F8 i.e. positive so the parabola opens upwards.

VERTEX
The vertex of the parabola X%5E2+=+4aY is the point (X=0,Y=0) with respect to the new co-ordinate system.
Now, x=8+X and y=2+Y.
So when X=0 and Y=0, x=8 and y=2.
So the co-ordinates of the vertex w.r.t. the old co-ordinate system are (x=8,y=2)

FOCUS
The co-ordinates of focus of the parabola X%5E2+=+4aY are (X=0,Y=a) in new co-ordinate system and so (x=8,y=2+a) in old co-ordinate system.
As here a=2 so the co-ordinates of the focus are (x=8,y=17%2F8).

DIRECTRIX
The equation of directrix of the parabola X%5E2+=+4aY is Y+a=0 in new co-ordinate system. Transferring to old co-ordinate system the equation is %28y-2%29%2B1%2F8=0 or y=15%2F8.

graph%28400%2C300%2C-1%2C19%2C-1%2C20%2C0.5%2A%28x-8%29%5E2%2B2%29