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]}
[Ans]

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