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Mathematics 14 Online
OpenStudy (marcoreus11):

Write an inequality for each graph help

OpenStudy (marcoreus11):

Help

OpenStudy (marcoreus11):

@jim_thompson5910 help

OpenStudy (3mar):

first of all, can you get the equation of the function graphed?

OpenStudy (marcoreus11):

arent we suppose to make an equation

OpenStudy (3mar):

I know

OpenStudy (3mar):

Get the equation of the graphed function, check which area/region represented in the graph, then apply the signs < > to match it. This is the sequence I know! Can you tell me a short one?

OpenStudy (marcoreus11):

can u just show me how this one is done cuz then i can use this as an example for the other problems i have

OpenStudy (marcoreus11):

right now i am just super confused

OpenStudy (3mar):

Please relax. You will get happy when you hit the target!

OpenStudy (marcoreus11):

okay

OpenStudy (3mar):

Do you know what the function graphed is?

OpenStudy (marcoreus11):

this is what i have y=-2lx-3l-4

OpenStudy (marcoreus11):

i think this is wrong

OpenStudy (3mar):

Let me see

OpenStudy (3mar):

May I share a useful "grapher" with you?

OpenStudy (marcoreus11):

sure

OpenStudy (3mar):

www.calorful.com

OpenStudy (3mar):

If you wish, test your functions, algebraic and graphically.

OpenStudy (marcoreus11):

can u just show me how this one is done

OpenStudy (3mar):

I am on it.

OpenStudy (3mar):

For the equation: it is not -| | as -| | is looking down, but the graph we have is looking up, so it will be +| |

OpenStudy (3mar):

Its head is at (1,-4)

OpenStudy (marcoreus11):

okay

OpenStudy (marcoreus11):

So what the whole thing look like?

OpenStudy (3mar):

Take that pic, please.

OpenStudy (marcoreus11):

?

OpenStudy (3mar):

Hope that is clear

OpenStudy (3mar):

Got it?

OpenStudy (marcoreus11):

so its only negative it faces down right?

OpenStudy (3mar):

Right. for this function and many others

OpenStudy (3mar):

Let's proceed?

OpenStudy (marcoreus11):

i guess

OpenStudy (danjs):

In general for an absolute value function f(x)... \[\large f(x) = a*\left| x-h \right|+k\] The vertex point of it is at the point (h,k) the slope of the line on the right half is 'a'

OpenStudy (danjs):

You have vertex at (h,k)=(1,-4), and slope a=2 \[y=2\left| x-1 \right|-4\]

OpenStudy (3mar):

Now we will test either area (inside the legs of the function and outer) to see which one is shaded. oK?

OpenStudy (marcoreus11):

i can't seem to understand the 2 where its coming from @DanJS

OpenStudy (danjs):

The function \[f(x)=\left| x \right|\] has a line slope 1 on the right... this one has slope a=2

OpenStudy (3mar):

It is a simple test. Let's pick the origin point as the test point. Simply the result of direct substitution lead you which sign you should choose.

OpenStudy (marcoreus11):

would the slope be 1/2 because u go 1 left up 2

OpenStudy (marcoreus11):

wouldnt?

OpenStudy (danjs):

\[f(x)=a*\left| x-h \right|+k\] slope is rise/run, change in y over change in x, here it is 2/1 = 2 For the general function above The 'a' is the slope , it squishes the graph horizontally or stretches it out.. the vertex point is at (h,k)

OpenStudy (marcoreus11):

oh i see where u r coming from

OpenStudy (danjs):

Like this one here, it is translated 2 left and 2 up, and has a slope of -1 on the right side. |dw:1475030585868:dw|

OpenStudy (3mar):

Can we finish this problem?

OpenStudy (3mar):

@MarcoReus11

OpenStudy (danjs):

For th example above... \[y=-1*\left| x-(-2) \right|+2\] \[y=-\left| x+2 \right|+2\]

OpenStudy (marcoreus11):

i think u mean by -2 for slope right?

OpenStudy (marcoreus11):

actually nevermind u r right my bad 2 divided by -2 is 1

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