Question

How do you find the vertex and intercepts for `y² - 4y - 8x +25 = 0`?

Answer 1

Vertex is `(21/8,2)`, x intercept is `x=25` and there are no y intercepts.

Explanation:

First reorder to get one of the variables (in this case x) on its own.`8x = y^2 - 4y +25` or `x = y^2/8 - y/2 +25/8`

Then use completing the squares to get the expression into the form `m(x+b/2)^2 + n`. This is easiest done before dividing across the board by 8.
`8x = (y-2)^2 -4 + 25 = (y-2)^2 +21`
`x = (y-2)^2/8 +21/8`

The vertex occurs where the bracketed term is zero and is therefore `(21/8,2)`
The x intercept occurs where `y=0` and is therefore `x=25`
The y intercepts occur where `x=0` and in this case it can be established by eye that the graph will never cross the y axis. Alternatively this can be shown using the quadratic formula.
`y = (-(-1/2) +- sqrt((1/2)^2 -4*1*(25/8)))/(2*1)`
`y = (1/2 +- sqrt(1/4 - 25/2))/2 = (1/2+- sqrt(2 - 25)/2)/2 = (1+-sqrt(-23))/4`
This shows that there are no real roots to the equation, only imaginary ones.