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Mathematics 6 Online
OpenStudy (anonymous):

for y=X^2-5 Do the following. Sketch a graph of the equation. Identify the vertex Compare the graph of y=f(x) of y(x^2) State any transformation used.

OpenStudy (anonymous):

to graph the equation u need the turning point, the x intercept and the y intercept

OpenStudy (anonymous):

looks just like \(y=x^2\) only shifted down 5 units

OpenStudy (anonymous):

i dont think she knows how that looks but if u do then there u have it and if u dont u can either do the working or look it up :)

OpenStudy (anonymous):

I do not know what it looks like

OpenStudy (anonymous):

okay, now to figure out the rest and how to get that darn version of the graph

OpenStudy (anonymous):

how do we find the vertex again?

OpenStudy (anonymous):

if i knew wat a vertex was then i would know wat to do :/ u c, in math I never learn off the names of things or these names ya'll use in USA is different from wat people in my country calls it :/

OpenStudy (anonymous):

which part of the curve is the vertex? the turning point?

OpenStudy (anonymous):

okay so the vertex is -5? therefore the bottom of the graph should start there not at 0?

OpenStudy (anonymous):

Yes I just looked at the picture.

OpenStudy (anonymous):

oh so the vertex is the turning point :P u should of just said that lol

OpenStudy (anonymous):

hell, I didn't even know that, it's asking for a ordered pair for the turning point?

OpenStudy (anonymous):

(0,-5)

OpenStudy (anonymous):

okay now to compare the two. gosh I hate freaking math,

OpenStudy (anonymous):

did the graph shift 5 units upwards, downwards, right, left?

OpenStudy (anonymous):

downwards

OpenStudy (anonymous):

okay so my graph should look like this. ? |dw:1364436616422:dw|

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