SOLUTION: Identify the vertex, axis of symmetry, and direction of opening for:
y=1/2(x-8)^2+2.
Thank you
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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) (Show Source): You can put this solution on YOUR website!
First re-arrange the given equation to compare with standard form.
or
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 .
This is a parabola with the Y-axis as its axis.
Comparing with the standard equation , a=.
AXIS OF SYMMETRY AND DIRECTION OF OPENING
Here the axis of symmetry is x = 8 and as i.e. positive so the parabola opens upwards.
VERTEX
The vertex of the parabola 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 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=).
DIRECTRIX
The equation of directrix of the parabola is Y+a=0 in new co-ordinate system. Transferring to old co-ordinate system the equation is or .
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