How do you find the vertex and intercepts for y^2=1/3x?

2 Answers
Dec 23, 2015

$\left(0 , 0\right)$ is only intercept.
$\text{vertex} = \left(\infty , \infty\right)$

Explanation:

For $y = 0$,
$\textcolor{w h i t e}{\times} x = 0$

Therefore $\left(0 , 0\right)$ is only intercept.

$y = \sqrt{\frac{1}{3} x}$
Therefore $y$ is positive for $\forall x \in {\mathbb{R}}^{+}$.

$y '$ is also positive for $\forall x \in {\mathbb{R}}^{+}$:
$y ' = \frac{1}{2 \sqrt{3} \sqrt{x}}$

Therefore there is no real vertex:
$\text{vertex} = \left(\infty , \infty\right)$

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

Dec 23, 2015

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

Explanation:

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

If $y = 0$ then $\frac{1}{3} x = 0$, so $x = 0$.

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

${y}^{2} \ge 0$ for any Real value of $y$, hence the minimum possible value of $x$ is $0$. So the vertex of the parabola is at $\left(0 , 0\right)$.

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]}