Question

How do you find the vertex and intercepts for `f(x)= x^2 + 3`?

Answer 1

Vertex and y - intercept at `(0,3)`. No `x` intercepts.

Explanation:

Using the discriminant of the quadratic formula:

`b^2-4ac` we see that:
`(0)^2-4(1)(3) = -12<0`
As the discriminant is less than `0` then the quadratic has no `x` intercepts.

To find the `y` intercept we can set`x=0` to get:

`f(0)=(0)^2+3=3`

To find the vertex, normally we could use the intercepts but this quadratic doesn't have any. We can use the vertex/ completed square form of the quadratic,

`f(x) = (x-a)^2+b`

Where the vertex is `(a,b)`

Note it is already in this form, i.e:

`(x-0)^2+3`
So reading off the values we get `(0,3)` for the vertex which coincidentally happens to be the `y` intercept as well.

Here is a graph of the parabola to better your understanding of what is going on, notice the parabola does not intercept the `x` axis and the turning point (vertex) as well as the `y` intercept are at `(0,3)`.

graph{x^2+3 [-9.25, 10.75, -1.12, 8.88]}