How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=x^2+4#?

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Answer 1 · 2017-05-15T08:59:23

#y# intercept is #4#, axis of symmetry is #x=0# i.e. #y#-axis and vertex is #(0,4)#.

This equation is of the form

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

whose axis of symmetry is #x-h=0# and vertex is #(h,k)#

As #y=f(x)=x^2+4=(x-0)^2+4#

Hence axis of symmetry is #x=0# i.e. #y#-axis

Vertex is #(0,4)#

To find #y# intercept we should have #x=0# and at #x=0#,

we have #y=0^2+4=4#

Hence #y# intercept is #4#.

graph{x^2+4 [-10, 10, 0, 10]}