How do you find the vertex and intercepts for #4y=x^2+4#?

1 Answer
Nov 27, 2015

So #P_("vertex") -> (x,y) -> (0,+1)#
The only intercept is on the y-axis and that is at the vertex.

Explanation:

Standard quadratic form:#color(white)(...)y= ax^2+bx+c#

Given: #4y=x^2+4#

Write as: #y=1/4x^2 +1#

Let any point on the graph be P

There is no #bx# term so the graph is symmetrical about the y-axis

Consequently #x=0#

The whole plot is transformed upwards from y=0 by the +1 and the amount of upward transformation is +1

So #P_("vertex") -> (x,y) -> (0,+1)#

Tony B