Question

What are the important points needed to graph `y= -x^2+2x+4`?

Answer 1

`x`-intercepts at `(1-sqrt5, 0)` and `(1+sqrt5, 0)`, `y`-intercept at `(0,4)` and a turning point at `(1,5)`.

Explanation:

So we have `y = -x^2 + 2x +4`, and usually the sorts of 'important' points that are standard for including on sketches of quadratics are axis intercepts and the turning points.

To find the `x`-intercept, simply let `y=0`, then:
`-x^2 + 2x +4 = 0`

Then we complete the square (this will also help with finding the turning point).

`x^2 - 2x + 1` is the perfect square, then we subtract one again to maintain the equality:
`-(x^2 - 2x + 1) + 1 +4 = 0`
`:. -(x-1)^2 + 5 = 0`

This is the 'turning-point' form of the quadratic, so you can read your stationary point right off: `(1,5)` (alternatively you could differentiate and solve `y' = 0`).

Now just transpose the equation:
`(x-1)^2 = 5`
`:. x- 1 = +- sqrt5`
`:. x = 1+-sqrt5`

The `y`-intercept is easy, When `x=0`, `y = 4`.

And there you have it!