# How do you find the vertex and intercepts for #x^2=8y#?

##### 2 Answers

#### Explanation:

#"the standard form of a parabola with the y-axis as it's"#

#"principal axis and opening vertically is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(x^2=4py)color(white)(2/2)|)))#

#"this parabola has it's vertex at the origin "(0,0)#

#• " if 4p ">0" then opens vertically up "uuu#

#• " if 4p "<0" then opens vertically down "nnn#

#x^2=8y" is in this form"#

#rArr4p=8#

#"since 4p ">0" then opens up"#

graph{x^2=8y [-10, 10, -5, 5]}

Vertex

#### Explanation:

The graph of

Since the coefficient of

Since

Hence, the vertex of

Since the vertex is the absolute minimum of

This result can be seen from the graph of

graph{x^2/8 [-16.11, 15.93, -2.15, 13.87]}