SOLUTION: The parabola with equation {{{y-3=(x-4)^2}}} has it's vertex at (4,3) and passes through (5,4). Find an equation of a different parabola with its vertex at (4,3)that passes through

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The parabola with equation {{{y-3=(x-4)^2}}} has it's vertex at (4,3) and passes through (5,4). Find an equation of a different parabola with its vertex at (4,3)that passes through      Log On


   



Question 138701: The parabola with equation y-3=%28x-4%29%5E2 has it's vertex at (4,3) and passes through (5,4). Find an equation of a different parabola with its vertex at (4,3)that passes through (5,4). Can't seem to figure this out. Do i just switch the x and y?
Thanks

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
The parabola with equation y-3=%28x-4%29%5E2 has it's vertex at (4,3) and passes through (5,4). Find an equation of a different parabola with its vertex at (4,3)that passes through (5,4). Can't seem to figure this out. Do i just switch the x and y?
Here's the graph of y-3=%28x-4%29%5E2 with the vertex (4,3) and 
the point (5,4).



That is a parabola which opens upward.  Let's sketch in one in
green that would have that same vertex(4,3) and pass through 
that same point (5,4). It will have to open to the right.



A parabola which open to the right or left has equation

%28y-k%29%5E2=4p%28x-h%29

So we substitute (h,k) = (4,3) 

%28y-3%29%5E2=4p%28x-4%29

Now we know that since (x,y) = (5,4) is a point on the
graph, that point must satisfy the equation: 

So
%284-3%29%5E2=4p%285-4%29

1%5E2=4p%281%29

1=4p

1%2F4=p 

Therefore the equartion is found by substituting p=1%2F4

into %28y-3%29%5E2=4p%28x-4%29

or

%28y-3%29%5E2=4%281%2F4%29%28x-4%29

%28y-3%29%5E2=1%28x-4%29

%28y-3%29%5E2=x-4

Edwin