# How do you find the vertex and intercepts for  y² = 128x?

Jun 9, 2016

The vertex is at $\left(0 , 0\right)$ and intercepts on both $x$-axis and

$y$-axis are zero.

#### Explanation:

It is apparent from the equation ${y}^{2} = 128 x$ that

$x > 0$ for all values of $y$ as $x = {y}^{2} / 128$ and

for each value of $x$, there are two values $y$ one $+ \sqrt{128 x}$ and other $= \sqrt{128 x}$,

except for $x = 0$, when $y = 0$ too.

Hence the vertex is at $\left(0 , 0\right)$ and clearly intercepts on $x$-axis and $y$-axis both are zero as putting either $x = 0$ or $y = 0$, both give other coordinate as $0$.

The parabola is symmetric around $x$-axis and is as given below.

graph{y^2=128x [-320, 960, -320, 320]}