Question

How do you use the important points to sketch the graph of `y= x^2 - 8x + 15`?

Answer 1

Here you go.

Explanation:

Basically, you need the `x`-intercepts (if any), the `y`-intercepts, and the vertex point. Knowing the concavity helps as well.

To get the `x`-intercept, let `y=0`.

`0=x^2-8x+15`

`(x-5)(x-3)=0`

So `x` can be `5` or `3` `->` the `x`-intercepts

For the `y`-intercept, I just let `x=0`. This gives the constant, so the `y`-intercept is at the point `(0,15)`.

For the vertex point, I can use `-b/(2a)`, so

`-b/(2a) = 8/2 = 4`

I'm not finished, I need to sub `x=4` back in.

`y=-1` when `x` is `4`.

Now I have the vertex `(4,-1)`, the `x`-intercepts `(3, 0)` and `(5, 0)`, the `y`-intercept `(0,15)`, and the concavity (concave up as it is a positive coefficient of `x^2`).

Now I graph.

graph{x^2-8x+15 [-10, 10, -5, 5]}