x-intercept and parabolas

[p_gorsky]

Hi, I was just wondering, when we are looking for the x-intercept and we are told we can set the equation equal to 0 to do this, is that always the case? Also, I understand pretty much everything about parabolas and what their points are etc but have no idea how you would go about finding the vertex. In the official GRE book on pg. 239 it shows it and one can clearly see it but no explanation of how to find it.

Thanks,
Pam

[gmbd97]

FOR FIRST QUESTION:
To get x intercept of any equation, we need to put y=0 , then solve the equation!! you are done.
Suppose, one equation is y=5x+5
so now, y=0, 0=5x+5
5x=-5
x=-1 (that is x intercept!)

FOR THE SECOND QUESTION:
In page 242 of ets official guide, 2nd edition , there is enough saying about the question you are asking!!

The graph of h(x)+c is the graph of h(x) shifted upward by c units.

that means, if the equation is y=(x-3)^2+2, the vertex will be 2 unit upward from x axis, here c=2

The graph of h(x)−c is the graph of h(x) shifted downward by c units.

that means,that means, if the equation is y=(x-3)^2-2, the vertex will be 2 unit downward from x axis, here c=2

The graph of h(x+c) is the graph of h(x) shifted to the left by c units.


that means,that means, if the equation is y=(x+3)^2-2, the vertex will be 3 unit left from y axis, here c=3

The graph of h(x−c) is the graph of h(x) shifted to the right by c units.

that means,that means, if the equation is y=(x-3)^2-2, the vertex will be 3 unit right from y axis, here c=3



So these are the basic things. Now for illustration, a equation is
y=(x+5)^2-3, then what is vertex?

answer: (-5,-3)

y=(x-5)^2-3, what is the vertex?

answer:(5,-3)

y=(x-5)^2+3, what is the vertex?

answer:(5,3)

y=(x+5)^2+3, what is the vertex?

answer:(-5,3)
HOPE it helped!!
GM

[tommywallach]

Wow! GM rocked this explanation so hard, I don't know what to do about it. I may have to go lie down or something! : )

-t