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

Jul 2, 2017

Vertex is at $\left(0 , 1\right)$, x-intercept is at $\left(\frac{1}{4} , 0\right)$ , y-intercept is at $\left(0 , 1\right)$.

#### Explanation:

${\left(y - 1\right)}^{2} = 4 x \mathmr{and} {\left(y - 1\right)}^{2} = 4 \left(x - 0\right)$.

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 $\left(0 , 1\right)$ . Putting $x = 0$ in the equation ,we can find

y-intercept as ${\left(y - 1\right)}^{2} = 4 \cdot 0 \mathmr{and} {\left(y - 1\right)}^{2} = 0 \mathmr{and} y - 1 = 0 \mathmr{and} y = 1$

So y-intercept is at $y = 1 \mathmr{and} \left(0 , 1\right)$. Putting $y = 0$ in the equation,

we can find x-intercept as ${\left(0 - 1\right)}^{2} = 4 \cdot x \mathmr{and} 4 x = 1 \mathmr{and} x = \frac{1}{4}$

So x-intercept is at $x = \frac{1}{4} \mathmr{and} \left(\frac{1}{4} , 0\right)$

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