Question

How do you graph `y= (x+1)^2 -10`?

Answer 1

Three points make a curve. You can solve for the vertex (minimum) and the (presumably) two x-intercepts, and sketch whatever connects all three points.

Normally you might do this:
`x = -b/(2a)`

But this form is easier. You basically have an `x^2` curve.

This one is shifted left 1 unit and down 10 units from `(0,0)`, because `x+1` in parentheses indicates a shift opposite to the sign (`+` is left, `-` is right), and the `10` outside of the parentheses corresponds to `+` as up and `-` as down.

So instead of the vertex at `(0,0)`, it is at `color(blue)(((-1,-10)))`. The x-intercepts, you get from solving the equation set to `y = 0`:

`0 = (x+1)^2 - 10`
`pmsqrt10 = x+1`

`color(blue)(x_"right") = sqrt10 - 1 color(blue)(~~ 2.162)`
`color(blue)(x_"left") = -sqrt10 - 1 color(blue)(~~ -4.162)`

Finally, if you want to, you get the y-intercept by setting `x = 0`:
`y = (0 + 1)^2 - 10 = -9`, and it is `color(blue)(((0, -9)))`.

You can see that here by clicking on the intercepts and the vertex:

graph{(x+1)^2 - 10 [-5, 5, -15, 5]}