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.
to graph the equation u need the turning point, the x intercept and the y intercept
looks just like \(y=x^2\) only shifted down 5 units
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 :)
I do not know what it looks like
okay, now to figure out the rest and how to get that darn version of the graph
how do we find the vertex again?
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 :/
which part of the curve is the vertex? the turning point?
okay so the vertex is -5? therefore the bottom of the graph should start there not at 0?
Yes I just looked at the picture.
oh so the vertex is the turning point :P u should of just said that lol
hell, I didn't even know that, it's asking for a ordered pair for the turning point?
(0,-5)
okay now to compare the two. gosh I hate freaking math,
did the graph shift 5 units upwards, downwards, right, left?
downwards
okay so my graph should look like this. ? |dw:1364436616422:dw|
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