How do you find the vertex and intercepts for #y = 1/20x^2#?

1 Answer
Alan P.
Jun 29, 2016

Vertex is at #(0,0)#
y-intercept at #0#
x-intercept (only one) at #0#

Explanation:

One way to look at this is to considered the standard vertex form:
#color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)#
with vertex at #(color(red)(a),color(blue)(b))#

the given equation: #y=1/20x^2# can be written in explicit standard vertex form as:
#color(white)("XXX")y=color(green)(1/20)(x-color(red)(0))^2+color(blue)(0)#
with vertex at #(color(red)(0),color(blue)(0))#

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")y=1/20(0)^2=0#

The x-intercept(s) is(are) the value(s) of #x# when #y=0#
#color(white)("XXX")0=1/20x^2 rarr x=0#

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