**the table represents a function. what is f(5)? –8 –1 1 8** This is a topic that many people are looking for. **g4site.com** is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, ** g4site.com ** would like to introduce to you **Finding the rule for a linear function when given a table – How to solve math problems**. Following along are instructions in the video below:

Right so given this table of values. What we need to do is we need need to figure out what is gonna be a linear function. So there a couple you need to remember about a function right a function has an input value.

Which is x and then were gonna have an output value so this f of x. That f. Represents.

The name of the function and x. Represents the value of the function at x. So what we did was i said well what about when my input is negative 1 my output on my function was a negative.

7. When i plug in a 0. I now get a 0 as my helpful so i need to think of so what am i doing to my function to get these values.

So the first thing.

I always like to look into is looking at adding and subtracting ok and lets say we have so we have f of x. Equals. X.

And the easiest thing to look at is heres my output right heres the input. So if i was going to say like lets say i plugged in 1 well. I have this then or im sorry if i plugged in like what is my input ok so thats my output value.

But here is the value of the input and now i probably got you to really confuse right this is really what we call just like the function. But its also calling the input particles. But if i plug in 1.

Which is my input. Im going to get out 1 for this function. But if i set like lets say f of x.

Equals.

X. Plus. 2.

And then i plug in a 1. I would get 1 plus. 2.

Which is equal to 3 ok. So thats how you could get something different if i had plugged in a one that i get out of 3. So what were going to do is i always like to look at what am i adding to subtracting first so i looked at this to go from negative.

7 to 2 negative 1 to negative. 7. I would have to subtract 8 and then what i see is this true for the rest of them well 0 8.

Does not give me 0.

And what 8 does not give me 7. So therefore subtracting or adding is not going to work so the next thing what about multiplication. If i multiply a number lets say for instance.

Lets say my new function was 2x well then two times one would still actually give me two in this example. But if i did half times three. Now i have two times three which will leave me six.

So i need to say well how could i can i multiply numbers to my x to give my f of x value of negative seven or to get my f of x value so if i plug in a negative one what do i have to do to get a negative seven. We already ruled out you cant add it subtract so what about multiplying. What do i have to multiply it by negative one to give me negative seven can you say well you can multiply by seven right so we could say f of x.

Is equal to 7 times x or 27 x. So geez. So we know that works for the first equation.

Does that work for the next one.

If i plug in a zero f of zero equals. 7 times zero do i get zero again well 7 times zero is obviously zero what about half of one f of 1 equals. 7 times 1.

Which equals. 7. Which again is my output.

And if i try f of two i get 7 times 2 well 7 times. 2 is 14. Which against my output.

So the first thing were doing one of these so therefore my linear function. Actually i should say is going to be f of x. Equals.

7 times x. Okay. Because what im doing is to get all my input.

Im multiplying by 7 to get my output alphabet to get my output guys all right so the main important thing youve always looked at to look for the relationships of first adding and subtracting if agnus objectives work then look to the relationship of multiplying or dividing and that doesnt work then theres another one you guys ill show you a little bit later of how you can combine the two of them till you find your linear function you .

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