How do you find the vertex and the intercepts for #f(x)=5x^2#?

1 Answer
Jun 3, 2017

Vertex: #(0,0)#
#x#-intercept: #y=0#
#y#-intercept: #x=0#

Explanation:

The vertex form of a quadratic equation is expressed as

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

Where #(h,k)# is the vertex. That vertex is a maximum if the coefficient of #x^2# term is negative. But the vertex is a minimum if the coefficient of the #x^2# term is positive.

For the equation we are given, we can easily add zeros to put it into vertex form

#y=5x^2# becomes

#y=5(x-0)^2+0#

This quadratic equation has a vertex at #(0,0)# and, because the coefficient #5# of the #x^2# term is positive, we know this vertex is a minimum.

The #y#-intercept occurs when #x=0#, which is again the origin at #(0,0)#. The #x#-intercept occurs when #y=0#, which can only happen when #x=0#. So the only #x# or #y# intercept happens at the vertex, all of which converge at the origin: #(0,0)#.

The graph is as follows:

graph{5x^2[-2,2,-1.5,10]}