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Mathematics 14 Online
OpenStudy (anonymous):

The graph above is a transformation of the function f(x)=|x| Write an equation for the function graphed above

OpenStudy (anonymous):

OpenStudy (anonymous):

@SolomonZelman

OpenStudy (anonymous):

@mathstudent55

OpenStudy (anonymous):

the function is a translation from the origin. we know because we know what the graph of the absolute value function looks like normally: http://www.wolframalpha.com/input/?i=%7Cx%7C

OpenStudy (anonymous):

also, it is dilated in some way

OpenStudy (anonymous):

let's deal with the translation first. the concave point at (-2, -2) has been translated from (0, 0). how can we do that using f(x) = |x|?

OpenStudy (anonymous):

@swagjason1998

OpenStudy (anonymous):

I know the function has been shifted to the left by 2 units

OpenStudy (anonymous):

It also is either compressed or stretched in some way

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

it's been shifted left and up by 2 units. to represent this as a function we use: f(x) = |x+2|+2 now,what about the dilation?

OpenStudy (anonymous):

I am not sure what you mean by dilation. Not familiar with that term

OpenStudy (anonymous):

@swagjason1998

OpenStudy (anonymous):

the function was scaled by some number. we have to find that number. when we do the function will be of the form f(x) = c|x+2|+2

OpenStudy (anonymous):

I see how to find the shift but how to find the scaling?

OpenStudy (anonymous):

in the function |x|, we note that the slope of the positive line is 1 find the slope of the new positive line

OpenStudy (anonymous):

|dw:1400462980194:dw|

OpenStudy (anonymous):

where do you see the slope of the positive line?

OpenStudy (anonymous):

refer to diagram

OpenStudy (anonymous):

the new slope is over 1 and up 1 if I am looking at the diagram

OpenStudy (anonymous):

wot

OpenStudy (anonymous):

not following you with the slope on a translated function

OpenStudy (anonymous):

I know what the abs(x) looks like on a graph

OpenStudy (anonymous):

sorry, my intuition for this kinda thing is weird

OpenStudy (anonymous):

so, in the function we have to write, what is the slope of the positive line?

OpenStudy (anonymous):

and the slope would come before the x

OpenStudy (anonymous):

slope of |x| = 1

OpenStudy (anonymous):

but we have |x+2| + 2

OpenStudy (anonymous):

@swagjason1998

OpenStudy (anonymous):

is that -1?

OpenStudy (anonymous):

@mathstudent55 @satellite73 @SolomonZelman

OpenStudy (anonymous):

n look at it & use slope formula

OpenStudy (anonymous):

what is n swag?

OpenStudy (anonymous):

n = no

OpenStudy (anonymous):

that is where I am lost

OpenStudy (anonymous):

so pick two points on the line and figure out the slope

OpenStudy (anonymous):

slope I got was 1/4

OpenStudy (anonymous):

@swagjason1998

OpenStudy (anonymous):

yep. scale is 1/4

OpenStudy (anonymous):

so now I put that into the translated function

OpenStudy (anonymous):

g(x) = 1/4x + 2 + 2

OpenStudy (anonymous):

|x+2|/4 + 2

OpenStudy (anonymous):

oh I was writing it |1/4x+2|+2

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