Find an y of a line y=4x+12

[Rabinov23]

Find an y of a line y=4x+12
Sorry I didn't provide workings as I DIDN'T get it and because I was out for few days from school because I was sick and so now I need to catch up the work that I missed out , so I missed this topic.
Any help will be appreciated

 

[lev888]

Can you post the exact question? As written it doesn't make sense. Did you read the corresponding chapter in the textbook?

 

[MarkFL]

What do you mean by "find an y?"

 

[Rabinov23]

sorry a

 

[Rabinov23]

not an

 

[pka]

Find an y of a line y=4x+12
Any help will be appreciated

Are you sure that you copied the question correctly?
\(\displaystyle (-2,4),~(0,12),~(-3,0)\) are all points on that line. The y value is the second in the pair.

 

[Rabinov23]

so is it 12

 

[Rabinov23]

but can you show me the way of doing it please

 

[MarkFL]

The value of \(y\) (the dependent variable as written) depends on the value of \(x\) (the independent variable).

 

[Rabinov23]

O sorry it is find the x and y intercept of the equation given

 

[pka]

but can you show me the way of doing it please

There is nothing to show. When \(\displaystyle x=0\) what does \(\displaystyle y=~?\)
When \(\displaystyle x=-3\) what does \(\displaystyle y=~?\)

 

[Rabinov23]

Is it not y=4x+12
so 0=4x+12
then -12/4 and 4x/4
-3=x

 

[Rabinov23]

I'm I right so now I need to find the y intercept

 

[pka]

Read your textbook. That is why you have a text book.
The When \(\displaystyle y-\)intercept is \(\displaystyle (0,b)\) and the \(\displaystyle x-\)intercept is \(\displaystyle (a,0)\)

 

[MarkFL]

I'm I right so now I need to find the y intercept

The \(y\)-axis is the line \(x=0\), so use that value for \(x\), and that will tell you where the \(y\)-intercept is.

 

[Rabinov23]

is y=12

 

[MarkFL]

is y=12

Yes:

[MATH]y=4(0)+12=12[/MATH]