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

1 Answer
Mar 2, 2016

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]}