Help with math?

[AniJan]

Can somone help me? I don't know what to do, nor do I understand problems like f(x)=-3x + 7 and h(x)= 1/2x +4. 

 

The concepts are hard..

[dalekphalm]

First:

 

This is homework from school, yes?

 

If so, dude you need a Private Math Tutor if you can't keep up with this kind of stuff.

 

The question is asking you to make a graph that represents the equation. A graphing Calc could do this no problem, of course - but I'm assuming you have to do it by hand and show your work.

[Guest]

f(x)=-3x+7 is literally the same as y=-3x+7

just plug x and y and then graph it out

 

 

[AniJan]

1 minute ago, dalekphalm said:

This is homework from school, yes?

 

Absolutely not. 7th grade isn't at this level yet. It's tutor homework.

[AniJan]

Just now, deXxterlab97 said:

f(x)=-3x+7 is literally the same as y=-3x+7

just plug x and y and then graph it out

 

 

huh.

wow.

[Guest]

Just now, AniJan said:

huh.

wow.

do you know how to do no 11 at least? 

 

[AniJan]

Just now, deXxterlab97 said:

do you know how to do no 11 at least? 

 

Of course.

The problem looked a bit different, so it kind of threw me off.

[dalekphalm]

4 minutes ago, AniJan said:

Absolutely not. 7th grade isn't at this level yet. It's tutor homework.

Err... so ask your Tutor for help...

 

The whole point of a tutor is to work with you on an individual level. You said right in the OP that you didn't "understand it".

 

That means you need to ask your tutor to explain it more or better. Or fire them and get a new/better Tutor.

[Digital]

What are you supposed to do here? Draw those functions in a coordinate system?

[AniJan]

3 minutes ago, dalekphalm said:

Err... so ask your Tutor for help...

 

The whole point of a tutor is to work with you on an individual level. You said right in the OP that you didn't "understand it".

 

That means you need to ask your tutor to explain it more or better. Or fire them and get a new/better Tutor.

Well, it's not his fault. Probably my fault for not taking deep notes.

 

And I guess the saying "I didn't understand it" was a bit of an exaggeration. That was entirely my fault for misleading you.

[dalekphalm]

Just now, AniJan said:

Well, it's not his fault. Probably my fault for not taking deep notes.

 

And I guess the saying "I didn't understand it" was a bit of an exaggeration. That was entirely my fault for misleading you.

Well, take this as a learning opportunity then. You either need to improve your note-taking skills, or he's not doing a proper job of tutoring you.

 

You need to find out which one is the problem.

 

Either way, talk to your Tutor - let him know you are having problems with these, and ask him to dedicate time to working on them with you.

[Atmos]

F(x) = y  in this scenario.

To graph a line written in the format "F(x) = m(x) + b"

is very, very simple once you understand the basics of it.

 

"m" is the slope of your line. So in example '12', "F(x) = -3x + 7" the slope of your line is "-3 " however, a slope must be a fraction, so we'll convert that to a fraction, which is easy as pi 

"-3" converted to a fraction is "-3 / 1" 

Now that you know what your slope is (Rise, over run) you know that the line will slope moving 3 units down your 'y axis', and one unit right along your 'x axis' from your 'y intercept'

 

The y intercept in this case is 'b' which we already know as "7" so draw out a graph and plot your first point on 7, then the next point three downwards and one to the right, following the slope of the line you already established earlier on. do this for one more plot point either from what you just graphed, or from your y-intercept again, then using a straight edge connect the three points continuing the line outwards with an arrow at each end and bang, you're done.

 

 

Once you get that, those kinds of problems are easy as heck. If the problem is not in "F(x) = mx + b" then you can very easily just manipulate the problem, adding and subtracting until you get it into such a format. h(x) and y both are equivalent in this regard to f(x), they're all the same concept in simplest terms.

[AniJan]

5 minutes ago, Atmos said:

F(x) = y  in this scenario.

To graph a line written in the format "F(x) = m(x) + b"

is very, very simple once you understand the basics of it.

 

"m" is the slope of your line. So in example '12', "F(x) = -3x + 7" the slope of your line is "-3 " however, a slope must be a fraction, so we'll convert that to a fraction, which is easy as pi 

"-3" converted to a fraction is "-3 / 1" 

Now that you know what your slope is (Rise, over run) you know that the line will slope moving 3 units down your 'y axis', and one unit right along your 'x axis' from your 'y intercept'

 

The y intercept in this case is 'b' which we already know as "7" so draw out a graph and plot your first point on 7, then the next point three downwards and one to the right, following the slope of the line you already established earlier on. do this for one more plot point either from what you just graphed, or from your y-intercept again, then using a straight edge connect the three points continuing the line outwards with an arrow at each end and bang, you're done.

 

 

Once you get that, those kinds of problems are easy as heck. If the problem is not in "F(x) = mx + b" then you can very easily just manipulate the problem, adding and subtracting until you get it into such a format. h(x) and y both are equivalent in this regard to f(x), they're all the same concept in simplest terms.

Thank you for the explanation.

 

However, in the second picture I posted, it displays a problem like f(x)= x squared -4. I was only exposed to the basics of graphing a quadratic equation like y= x squared. What do you with the -4?

[anothertom]

11 minutes ago, AniJan said:

However, in the second picture I posted, it displays a problem like f(x)= x squared -4. I was only exposed to the basics of graphing a quadratic equation like y= x squared. What do you with the -4?

So you've been taught a very restricted form of this topic, not that it's your fault it's just how school works. [Feel free to skip to the second paragraph.]

Any equation dependent on a single variable, which in this case is X, can be expressed in the form of f(x)= a + bX¹ + cX^-1 + dX² +eX^-2... ... ... Where any letter in lower case is a constant, X is the variable and f(x) in word form means 'function of X' (and is plotted on the Y axis). If you haven't reached negative exponents yet then X^-n == 1/(Xⁿ) but don't worry if that went over your head. What you've been taught is this equation but only the first few parts of it.

 

If you're only instruction is to plot the graph then you can treat the equation exactly the way you would a linear graph (where there are no terms above X¹), draw a table with two rows, and 11 (narrow) columns, X in the top, Y in the bottom, along the X row number the columns -5 to +5, then put each X value into the function and write the result in the Y row. This is f(x) and gives the Y values so you can then plot the graph.

 

Because you're presumably plotting these by hand, you'll quickly get numbers which are impractical to draw out without using large amounts of paper so unless you've been given a set range for the X and Y axes, there's not much point going outside -4 to +4 on the X and +20 and -20 on the Y (and you don't have to have the same scale on both axes).

 

If you're struggling with the work your tutor is giving you then you need to tell them, it is completely their job to make sure you are understanding the topic and to provide assistance, that's why they're being paid.

[Atmos]

51 minutes ago, AniJan said:

Thank you for the explanation.

 

However, in the second picture I posted, it displays a problem like f(x)= x squared -4. I was only exposed to the basics of graphing a quadratic equation like y= x squared. What do you with the -4?

Wonderful that your tutor starts you out on basic slope intercept and point slope form equations, then straight to Parabolas and quadratics lol.

 

Graphing a parabola that is given in form "F(x) = a(x-h)^2 + k" where (h, k) are the vertex of your parabola. sounds somewhat complex at the start of it, but luckily those examples are pretty simple and you can ignore the variable "h" which only serves to complicate things a bit. 'H' represents your horizontal shift along the 'x axis', and 'K' represents the vertical shift along your 'y axis'.

 

In example '30' your equation is written as "F(x) = 3x^2 + 4" which is in format "F(x) = a(x-0)^2 + k"

So, we know that there is no horizontal shift from 'y = 0' and we know that we'll need to move the vertex of the parabola four units upwards from 'x = 0' because we have a value of 'k = 4'.

 

Now we also know that the parabola is going to have vertical stretch thanks to that value of 'a = 3'

We can easily find points to graph the parabola by constructing a table at this point using the data we already have. We know that f(x) or 'y' is equal to 3(x)^2 so all we need to do is plug in a few numbers for x

'x' values  |  'y' values
-------------------------
       0    |   3(0)^2   
       1    |   3(1)^2
       2    |   3(2)^2

y1 = 0
y2 = 3
y3 = 12

So now we know the vertex of the parabola, and the first two points along the right side of the parabola.

Our vertex is at (0,4) along the y-axis - which is just y1+4

The first point on the right side is (1,7)  - which is just y2+4

The second point in the right side is at (2,16) - which is just y3+4

Now you can use that axis of symmetry to equate the two points on the left side as well, being respectively, (-1,7) and (-2,16)

 

Now, that's a lot to take in, which surprises me why a tutor would drop that on you, especially given you're still not that comfortable with even point slope.

 

If you're concerned about whether or not you got a parabola correct, then you can use one of the plethora of online parabola graphing tools, however, and this is very important. Actually do it yourself and graph it out before checking it on the tool, don't just use the tool to solve the problems or else you'll never learn how to do it yourself

The one for reference I've used for a long time now is https://www.desmos.com/calculator

[Guest]

40 minutes ago, AniJan said:

Thank you for the explanation.

 

However, in the second picture I posted, it displays a problem like f(x)= x squared -4. I was only exposed to the basics of graphing a quadratic equation like y= x squared. What do you with the -4?

Y=x2-4 then you move up by 4 iirc

[AniJan]

20 minutes ago, anothertom said:

it's just how school works

I'm in 7th grade. This is tutor homework.

 

 

EDIT: Nevermind. You're already aware. Sorry.

[anothertom]

20 minutes ago, anothertom said:

 it's just how school works.

2 minutes ago, AniJan said:

I'm in 7th grade. This is tutor homework.

For 'school' read 'education'. You get lied to progressively but slowly approach something nearer what's actually true.

[AniJan]

14 minutes ago, Atmos said:

you're still not that comfortable with even point slope.

Oh, I'm comfortable with point slope form.

I was just confused with the "f(x)" because it seemed like a completely different format to me than the regular one.

But now I know that f(x) is basically the same thing as y. ty and everyone else who helped me.

[AniJan]

20 minutes ago, deXxterlab97 said:

Y=x2-4 then you move up by 4 iirc

Wait, so you just bring the whole curve up with you?

[Guest]

3 minutes ago, AniJan said:

Wait, so you just bring the whole curve up with you?

Either up or down, been a while since I do that. Just make an y=kx and y=kx-4 and compare if they move up or down

[WindirBear]

 

[yathis]

9 hours ago, AniJan said:

Can somone help me? I don't know what to do, nor do I understand problems like f(x)=-3x + 7 and h(x)= 1/2x +4. 

 

The concepts are hard..

Not really, pretty simple, besides the fx shit. Function of X is what its called no?

There are some really great math websites on the web you could have easily looked at for your homework, but you want it done for you.

[AniJan]

12 hours ago, yathis said:

Not really, pretty simple, besides the fx shit. Function of X is what its called no?

There are some really great math websites on the web you could have easily looked at for your homework, but you want it done for you.

No, not at all. I was simply perplexed on some problems, hence I asked the community here since there are tons of people that are experienced.

Is there a problem with doing such thing?

Don't jump to conclusions.

[yathis]

It is your homework no?