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

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Answer 1 · 2015-11-27T14:42:22

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

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

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