How do you find the vertex and the intercepts for #f(x)=x^2-4#?

1 Answer
Apr 8, 2016

vertex: #(0,-4)#
x-intercepts: #+-2#
y-intercept: #(-4)#

Explanation:

General vertex form
#color(white)("XXX")f(x)=color(green)(m)(x-a)^2+b# with vertex at #(a,b)#

Writing given equation: #f(x)=x^2-4# in vertex form:
#color(white)("XXX")f(x)=1(x-0)^2+(-4)# with vertex at #(0,-4)#

x-intercepts are the possible values of #x# when #f(x)=0#
#color(white)("XXX")0=x^2-4#
#color(white)("XXX")rarr x=+-2#

y-intercept is the value of #y(=f(x))# when #x=0#
#color(white)("XXX")y=f(x=0) = 0^2-4=-4#

graph{x^2-4 [-5.717, 6.773, -4.885, 1.355]}