What is the vertex form of $x = {(y - 3)}^{2} + 41$ $x = (y - 3)^2 + 41$ ?

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Answer 1 · 2017-08-03T15:03:44

$x = {(y - 3)}^{2} + 41$ $x = (y - 3)^2 + 41$ is in the vertex form.

The vertex form for a parabola that opens to the left or right is:

$x = \frac{1}{4 f} {(y - k)}^{2} + h [1]$ $x = 1/(4f)(y-k)^2+h" [1]"$

where $(h, k)$ $(h,k)$ is the vertex and $f = y_{focus} - k$ $f = y_"focus"-k$ .

The given equation

$x = {(y - 3)}^{2} + 41$ $x = (y - 3)^2 + 41$

is already in the form of equation [1] where $(h, k) = (41, 3) and f = \frac{1}{4}$ $(h,k) = (41,3) and f = 1/4$ .

What is the vertex form of x=(y−3)2+41x = (y - 3)^2 + 41?