Question

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

Answer 1

The vertex is at `(0,0)` and intercepts on both `x`-axis and

`y`-axis are zero.

Explanation:

It is apparent from the equation `y^2=128x` 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(128x)` and other `=sqrt(128x)`,

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

Hence the vertex is at `(0,0)` 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]}