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How do you find the vertex and intercepts for $y^{2} = \frac{1}{3} x$ $y^2=1/3x$ ?

2 Answers

Dec 23, 2015

$(0, 0)$ $(0,0)$ is only intercept.
$vertex = (\infty, \infty)$ $"vertex"=(oo,oo)$

Explanation:

For $y = 0$ $y=0$ ,
$\times x = 0$ $color(white)xxx=0$

Therefore $(0, 0)$ $(0,0)$ is only intercept.

$y = \sqrt{\frac{1}{3} x}$ $y=sqrt(1/3x)$
Therefore $y$ $y$ is positive for $AA x inRR^+$ .

$y'$ is also positive for $AA x inRR^+$ :
$y'=1/(2sqrt3sqrtx)$

Therefore there is no real vertex:
$"vertex"=(oo,oo)$

graph{sqrt(1/3x) [-10, 10, -5, 5]}

Dec 23, 2015

The vertex and intercepts are all at $(0, 0)$

Explanation:

If $x = 0$ then $y^2 = 0$ , so $y = 0$ .

If $y = 0$ then $1/3 x = 0$ , so $x = 0$ .

So the only $x$ and $y$ intercepts are at $(0, 0)$

$y^2 >= 0$ for any Real value of $y$ , hence the minimum possible value of $x$ is $0$ . So the vertex of the parabola is at $(0, 0)$ .

Both positive and negative values are possible for $y$ .

$y$ is not a function of $x$ , but $x$ is a function of $y$ ...

graph{y^2=x/3 [-5.22, 12.56, -4.484, 4.405]}

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