SOLUTION: how do i graph the parabola (y-9)^2=12(x-8)

Question 1081057: how do i graph the parabola (y-9)^2=12(x-8)
Found 2 solutions by josgarithmetic, KMST:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
That form of equation is what could happen if you used the definition given directrix and focus, and derived the equation. Yours is for parabola with horizontal symmetry axis, concave to the right.

Vertex is read from the equation: (8,9).

Find y-intercepts: Should be none.

----you can see this will not be real numbers.

Find x-intercept:

Point (14.75, 0).

Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
That is a modified form of the parabola .
You probably know that the graph of looks like this:
.
Nice graph, symmetrical about the positive y-axis (the line),
with a vertex at point (0,0), the origin.
Each point has a twin that is its reflection about the y-axis,
except the vertex.
The vertex is a unique point, and its reflection is the vertex itself.

If you swap the x for the y and the y for the x, you get :
.
Nice graph, symmetrical about the positive x-axis,
with a vertex at point (0,0), the origin.
Points with <--> , and ,
have .
Points with <--> , and ,
have .

If you graphed them on wax paper (or anything translucent enough),
flipping the paper so as to swap the arrows for the x- and y-axes
would turn one graph into the other.

is a version of ,
with a vertex at (8,9) ,
the unique point with <---> .
Points with <--> , and ,
have <---> .
Points with <--> , and ,
have <---> .

So, the graph is like the graph of ,
except it has been moved up and right to shift the vertex to (8,9),
its axis of symmetry is now the line ,
and it has been shrunk along the x-direction by a factor of ,
so you go much farther from the vertex in the x-direction
and get the same distance away from its axis of symmetry:
,
.
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