Question

How to graph a parabola `y=x^2-4x-5`?

Answer 1

You can start by finding the x-intercepts, if there are any. That is when y = 0.

`0 = x^2 - 4x - 5`
`0 = (x - 5)(x + 1)`

So you have one intercept at `x = 5` and one at `x = -1`. Or, at `"(5, 0)"` and `"(-1, 0)"`.

Then, you can find where the vertex is (where the graph turns around). This can be found using:

`x = (-b)/(2a)`

`(-(-4))/(2*1) = 4/2 = 2`

Then you can plug it into the original equation and solve for the y-coordinate.

Plugging it in:
`x^2 - 4x - 5 = y`
`2^2 - 4*2 - 5 = 4 - 8 - 5 = -9`

So I would expect a vertex at `"(2, -9)"`.

If you want to go even further, you can solve to see if there are any y-intercepts (at `x = 0`).

`y = 0^2 - 4*0 - 5 = -5`

So there is also a y-intercept at `"(0, -5)"`.

Here's the graph:
graph{x^2 - 4x - 5 [-20.07, 20.06, -10.03, 10.04]}

You can click on the graphed curve to locate the intercepts and vertex.