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First, use the given infiltration to construct the equation of line A, then set y equals to zero and solve for x. The x's that you solved for are the x-intercepts.
You can use the point slope formula and plug all the given info in. Do you remember the point slope formula?
I have a problem understanding where to put the numbers... In what cases do you replace it for Y and in what cases do u replace it for b?
\(\Large y-y_1=m(x-x_1)\) Where x1,y1 is the given point and m is the slope
That is the point slope formula ^^
is there any way to solve the problem without the point slope formula? because the solution doesnt use it
Or you can use y=mx+b Plug x,y by the given point and m for the slope Then solve for b After solving for b, you then go back and write y=mx+b again and plug in the y-intercept (b) and slope. Then, you have the equation solved.
yea thats what i have a problem understanding... in what case do you use it for b and in what case do you use it for Y???????????
All the matters is to construct the equation of line A so you can set y to zero and solve for the x-intercept. There's two ways you can do it.
i really need to understand that for future problems
Yes, i understand. |dw:1439228871702:dw|
then what about the Y what is the y then?
m and b (the slope and the y-intercept) are constants. Meaning, when writing the equation, the numbers are giving for them. But x and y is changing. It does not have a constant number. the x is the input and any numbers from the x-axis can be plugged into it. There's no constant, specific number to plug. It is changing. Same goes to y. Y is changing and it is the output. You can plug whatever number you want from the y-axis. There's no specific number.
For example. The graph y=x |dw:1439229116348:dw|
so when it says (2,7) is on the number line does that mean it's a constant?
But when you have a slope and the y-intercept (m and b) then the line changes position or shifts from y=x.
So can you tell me how to spot the constant?
That means that line A passes through this point (2,7) as well as many many other points. But this point can locate where the line is. You can think of it as a pin point on the graph and you can draw a line that passes through it. Of course, you can't draw a line from nowhere, you have to be given an info about the slope so we can understand what is the steepness of the line, does it have any or it does not. But it has, and the slope is already given to you. It is -5/3.
The only constants are the slope (m) and the y-intercept (b). A point is not a constant on the equation because (x,y) keeps changing. (2,7) is just a point that we are giving the info of that the line passes through.
Makes sense? Can we do some math now?
oh okay so do we use (-2,7) to find the constant?
oh okay and 7 is the b intercept right?
Nopes. 7 is a y-value that the pin point is located on. (2,7) where x=2 and y=7 is just a pin point ion the graph that gives us a sense where the line is positioned.
Don't think of y-interpret as any y-value it is a y-value where the line crosses the y-intercept. A y-intercept will always have x as 0. If (2,7) was (0,7) then (0,7) would be the y-intercept. An example of y-intercept i'm going to show on a made up line.|dw:1439229765311:dw|
oh okay i got it
Good. Can you construct the equation now?
oh okay so use the 2
y=mx+b there's an x and a y in this formula. We can't just discard the x like that. We have to use the point (2,7) and plug in for x and y in here. Also, we can't discard the 2 in (2,7) :)
so whats the next step?
Now, we have the two constants m=-5/3 and b=31/3 So, y=mx+b the equation is y=(-5/3)+31/3|dw:1439231431455:dw|
The next step is to find the x-intercept. How would you do that?
no we did it wrong
it's suppoed to be -5/3(-2) not positive 2
so it's -5/3+11/3
Correct. when you said "so when it says (2,7) is on the number line does that mean it's a constant?" i thought it was (2,7), i didn't go back and check what it was originally.
how come this works?
pardon my mouse handwriting
it's ok i get it now thanks you