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

Mar 2, 2016

Vertex and y - intercept at $\left(0 , 3\right)$. No $x$ intercepts.

#### Explanation:

Using the discriminant of the quadratic formula:

${b}^{2} - 4 a c$ we see that:
${\left(0\right)}^{2} - 4 \left(1\right) \left(3\right) = - 12 < 0$
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:

$f \left(0\right) = {\left(0\right)}^{2} + 3 = 3$

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,

$f \left(x\right) = {\left(x - a\right)}^{2} + b$

Where the vertex is $\left(a , b\right)$

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

${\left(x - 0\right)}^{2} + 3$
So reading off the values we get $\left(0 , 3\right)$ 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 $\left(0 , 3\right)$.

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