How do you find the vertex and intercepts for #(y - 1)^2 = 4x#?

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Answer 1 · 2017-07-02T11:02:58

Vertex is at #(0,1)#, x-intercept is at # (1/4,0) # , y-intercept is at # (0,1)#.

#(y-1)^2=4x or (y-1)^2=4(x-0)#.

This is parabola opening right. Comparing with standard equation of

#(y-k)^2=4a(x-h) ; (h,k)# being vertex , we get here #h=0,k=1#.

So vertex is at #(0,1)# . Putting #x=0# in the equation ,we can find

y-intercept as #(y-1)^2 = 4*0 or (y-1)^2=0 or y-1= 0 or y=1#

So y-intercept is at #y=1 or (0,1)#. Putting #y=0# in the equation,

we can find x-intercept as #(0-1)^2 = 4*x or 4x= 1 or x=1/4#

So x-intercept is at #x=1/4 or (1/4,0) #

graph{(y-1)^2=4x [-20, 20, -10, 10]}
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