f(x) = x^2 - 3x find f(-8)

P(T)=4t -5 Find P(T-2)

G(A)=3^3a-2: Find G(1)

G(X) =3x -3 : Find G(-6)

H(T) =2times 5^-t-1. Find H(-2)

G(A)=4a: Find G(2a)

f(x) = x^2 - 3x find f(-8)

P(T)=4t -5 Find P(T-2)

G(A)=3^3a-2: Find G(1)

G(X) =3x -3 : Find G(-6)

H(T) =2times 5^-t-1. Find H(-2)

G(A)=4a: Find G(2a)

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1. If given f(x) = x^{2} - 3x then find f(-8).

Substituting -8 for x we get (-8)^{2} - 3(-8), which gives us 88.

2. If given p(t) = 4t - 5 then find p(t - 2).

Substituting t - 2 for t we get 4(t - 2) - 5, which gives us 8t -13.

3. If given w(n) = 4n + 2 then find w(3n).

Substituting 3n for n we get 4(3n) + 2, which gives us 12n + 2.

4. If given g(a) = 3^{3a-2} then find g(1).

Substituting 1 for a we get 3^{3(1) - 2}, which gives us 3^{1} or simply 3.

5. If given g(x) = 3x - 3 then find g(-6).

Substituting -6 for x we get 3(-6) - 3, which gives us -21.

Here's another explanation of what the heck functions even are that will hopefully make sense to you.

Okay, so the first thing to remember is that we're dealing with graphs here. Your basic algebraic equation like y = mx + b is the equation of a straight line. No matter what's on the right side of the equation, though, it corresponds to some sort of graph, whether that be a parabola or anything else. Basically, what we're saying when we change the "y" to "f(x)" is that the x and y values of any given point on that line will share the same relationship to each other no matter what. For any x value you want to work with in this equation, you can use the same exact rule to determine the corresponding y value.

For example, y = 3x means that for any value you pick for x, the y value will be three times that much. The term "f(x)" actually means "the function of x" or to put it another way, "the value of y in terms of x". The whole "f" thing doesn't really matter, it can be any placeholder letter you want and still make sense. That's why some of your problems involved things like "W(N)" and "G(A)". The actual letters don't matter at all; it's just a math way of saying that the y value will be written in terms of the variable inside the parentheses.

You can think of functions as a little machine on your page. If your machine is "f(x) = 3x + 4", then that means that if you start with an x and you want to end up with the y that goes with it, you can just plug the x into the 3x + 4 machine and you'll come out with the matching y.

Let's take the first problem as an example. Your first machine is f(x) = x^2 - 3x, and they want you to find f(-8). That just means they've handed you the value "-8" for your starting x, and they want you to put it in the machine and see what comes out. So you substitute -8 for any x values in the machine's rule, which is x^2 - 3x. You'll end up with:

(-8)^2 - 3(-8)

which will work out to

64 - (-24)

64 + 24

88

So you've used the function machine to find the corresponding y value for when x is -8. That's it!

I hope this helped!

Answering this one also answers your other question. Simply substitute -8 wherever there is an x and simplify.

f(-8) = (-8)^{2} - 3(-8)

-8^{2} is 64, and -3 * -8 is 24, so your answer is:

f(-8) = 88

With this information, you should be able to work your other problem on your own. :)

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## Comments

Thank you, but I still don't understand. I am about to give up!

I'm assuming you're encountering this problem in an algebra class? The first time you see functions, they don't make any sense, and they're probably not going to for some time. You may find it easier for now to just accept that there are things here that won't make sense, and it's ok for them not to. Algebra is a skills class, you don't learn anything useful about the universe. Now, when you go and take a physics class, you do more intense algebra than you've ever done AND you learn about the world. But here, now, this thing doesn't have to make sense.

The only thing that has to make sense right now is this: f(x) means that for every place you see an x, you can put anything you want instead of the x. Here are some examples:

f(x) = x + 1

f(y) = y + 1 (replace with a y)

f(anything) = anything + 1

f(dinner) = dinner + 1

See? So when it's a number, like you have here, you just replace with the number. After that, a simple computation will turn it into a single number, and you have a value.

So, f(-8) = (-8)

^{2}- 3(-8)Not too long ago I'm sure you did some linear equations and parabolas (the dreaded quadratic, complete with the quadratic equation). So you looked at something like this:

y = 5x + 6 (linear equation)

You recall, hopefully, that that represents a line with a y-intercept at 6 and a slope of 5? Well, for a function, just replace y with f(x), like so:

f(x) = 5x + 6

Previously, you would make a chart like this to graph the line:

x | y

0 | 6

1 | 11

-6 | 0

Now, after you've changed y to f(x), you'll do the same thing, like so:

x | f(x)

0 | 6

1 | 11

-6 | 0

Changing the notation looks like it's just a bunch of silliness, and I have to admit the first time I encountered functions, I thought they were just silly. And all the words around them just make it hopelessly complicated for no good reason, especially considering that linear equations worked fine without all that jargon, right?

In calculus, you'll study functions in depth and learn very rigorous tools and methods to study how they change. For every calculus student, there's an "aha!" moment when they feel like they've learned something fundamental about how the universe works.

But that happens in calculus. By then, you'll have studied functions in algebra, trigonometry, and precalculus. That's what it takes for them to finally make sense. So, they're introduced in the curriculum here so you can start to learn how to use algebra to work with them, but full understanding isn't available for a little while longer.

That's one of the things that makes algebra so tough: there are quite a few topics that don't make any sense until you see them in a different context in a different class.

So keep your head up!

Does this help?