Question

How do you use the important points to sketch the graph of `y=x^2+2x+3`?

Answer 1

To sketch the graph, you will need the vertex, axis of symmetry and `x` and `y`-intercepts.

Explanation:

First, you turn `x^2 + 2x + 3` into a complet square as follows:

`y = x^2 + 2x + (1)^2 + 3 -(1)^2`

`y = (x+1)^2 +3-1`

`y = (x+1)^2 +2`

Knowing this and comparing it to the vertex form of a parabola

`y = a(x-h)^2 + k`

you can determine the vertex by using the point `(h,k)`, which is `(-1,2)`.

So `V(-1,2)` and the axis of symmetry `= -1` since it equals the `h` value of the vertex.

When `x=0`, you can find the `y`-intercept, and when `y=0`, you can find the `x`-intercept.

Therefore,

`y = (0+1)^2 +2`

`y`-intercept `= 3`

and

`0 = (x+1)^2 +2`

Rearranging,

`(x+1)^2 = -2`

`x+1 = +-sqrt(-2)`

`x = -1 +- sqrt(-2)`

Since `sqrt(-2) !in RR`, you can say that this parabola has no `x` intercepts, i.e. it doesn't cross the `x` axis.

Therefore, we can sketch our graph with the knowledge that

• vertex = `(-1,2)`
• axis of symmetry `= -1`
• `y`-intercept `= 3`
• `x`-intercept `= O/` [no `x`-intercept(s)]

graph{x^2+2x+3 [-5, 5, -10, 10]}

Hope this helped!