Why are the harder examples always easier than the practice problems?

yeah ive struggled with the fact that the"harder" examples is just one example and there are multiple kinds of problems it does not show you how to solve

Is there an algebraic way of solving this kind of problem? I'm aware that you can simply plug in the possible values for the variables to reach an answer, but I'm trying to fully understand the topic as well. EDIT: I've looked up ways to solve this, and it seems like you have to graph each inequality. Are graphing and plugging in points the only two options when trying to solve this problem?

Unfortunately, I don't think so. The problem is that the `x` is in the exponent section, and although it _might_ be solved using logarithms or other advanced math, it's probably easier and faster to just plug in the numbers.

Is there another way to find out the answer other than plugging in random numbers?

I'd say plugging in random numbers is the most accurate solution you could do here, because just doing brute force math without much logic and reasoning lowers the chance for you to make a mistake. However, you're right, you can solve this problem another, faster way. You're asked to find the graph of the negative of the function you're given. Putting a negative on a function like that basically means you're flipping it about the x-axis. So if we know the shape of the original function, we can just flip it horizontally and get our answer from there. The function's equation looks exponential. The value of the constant being raised to a variable power is 1/3 here, which tells us that we're looking at exponential decay here and not growth (it's less than 1). We also have a minus 1 tacked onto the end. The minus 1 moves the whole function down by 1. So normally, the function would start at infinity, and decay exponentially until it hits an asymptote of -1. We then flip this in our heads (or draw a sketch on the paper and flip that). Now our function has an asymptote at positive 1, and decays exponentially but in the opposite direction, getting more positive by decreasing amounts instead of getting more negative in decreasing amounts. Choice D) is the only one that matches this.

I didn’t get why he took -f(x) can some please explain. Can’t we just do it f(x) without the minus

The question is asking to find the graph of -f(x).

What if x goes -1......-2......-3.....-4 and so on?

When you say a function increases, then you're assuming x is increasing unless otherwise specified.

I need all of you to solve this: 2 - X = X^2 + 4

2 - x = x^2 + 4 x^2 + 4 + x - 2 = 0 x^2 + x + 2 = 0 (x + 2)(x - 1) = 0 x = -2 and 1

Main content

Course: Digital SAT Math > Unit 12

Lesson 12: Exponential graphs: advanced

Nonlinear equation graphs — Harder example

Name: Nonlinear equation graphs — Harder example
Uploaded: 2018-12-21T15:24:59Z
Description: Watch Sal work through a harder Nonlinear equation graphs problem.

Google Classroom

Watch Sal work through a harder Nonlinear equation graphs problem.

Want to join the conversation?

Sort by:

Liam Russell
Posted 3 years ago. Direct link to Liam Russell's post “Why are the harder exampl...”
Why are the harder examples always easier than the practice problems?
Button navigates to signup pageComment on Liam Russell's post “Why are the harder exampl...”
(27 votes)
Answer
- bella
  Posted 2 years ago. Direct link to bella's post “yeah ive struggled with t...”
  yeah ive struggled with the fact that the"harder" examples is just one example and there are multiple kinds of problems it does not show you how to solve
  Button navigates to signup page
  (13 votes)
Gwynn Major
Posted 5 years ago. Direct link to Gwynn Major's post “Is there an algebraic way...”
Is there an algebraic way of solving this kind of problem? I'm aware that you can simply plug in the possible values for the variables to reach an answer, but I'm trying to fully understand the topic as well.

EDIT: I've looked up ways to solve this, and it seems like you have to graph each inequality. Are graphing and plugging in points the only two options when trying to solve this problem?
Button navigates to signup pageButton navigates to signup page
(13 votes)
Answer
- Lin Gh.
  Posted 5 years ago. Direct link to Lin Gh.'s post “Unfortunately, I don't th...”
  Unfortunately, I don't think so. The problem is that the x is in the exponent section, and although it might be solved using logarithms or other advanced math, it's probably easier and faster to just plug in the numbers.
  Comment on Lin Gh.'s post “Unfortunately, I don't th...”
  (11 votes)
muadhmamun782
Posted 7 months ago. Direct link to muadhmamun782's post “lets go October sat here ...”
lets go October sat here baby!! Need a thick 800 math score
Button navigates to signup pageButton navigates to signup page
(17 votes)
Answer
arodriguez24
Posted a year ago. Direct link to arodriguez24's post “Is there another way to f...”
Is there another way to find out the answer other than plugging in random numbers?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Hecretary Bird
  Posted a year ago. Direct link to Hecretary Bird's post “I'd say plugging in rando...”
  I'd say plugging in random numbers is the most accurate solution you could do here, because just doing brute force math without much logic and reasoning lowers the chance for you to make a mistake. However, you're right, you can solve this problem another, faster way.
  
  You're asked to find the graph of the negative of the function you're given. Putting a negative on a function like that basically means you're flipping it about the x-axis. So if we know the shape of the original function, we can just flip it horizontally and get our answer from there.
  The function's equation looks exponential. The value of the constant being raised to a variable power is 1/3 here, which tells us that we're looking at exponential decay here and not growth (it's less than 1). We also have a minus 1 tacked onto the end. The minus 1 moves the whole function down by 1. So normally, the function would start at infinity, and decay exponentially until it hits an asymptote of -1.
  We then flip this in our heads (or draw a sketch on the paper and flip that). Now our function has an asymptote at positive 1, and decays exponentially but in the opposite direction, getting more positive by decreasing amounts instead of getting more negative in decreasing amounts. Choice D) is the only one that matches this.
  Comment on Hecretary Bird's post “I'd say plugging in rando...”
  (12 votes)
talinek
Posted a year ago. Direct link to talinek's post “I didn’t get why he took ...”
I didn’t get why he took -f(x) can some please explain. Can’t we just do it f(x) without the minus
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- SP
  Posted a year ago. Direct link to SP's post “The question is asking to...”
  The question is asking to find the graph of -f(x).
  Comment on SP's post “The question is asking to...”
  (7 votes)
khalifaha482
Posted 8 months ago. Direct link to khalifaha482's post “the harder example is eas...”
the harder example is easier than the basic one
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
Sameer .K. Banthia
Posted 5 years ago. Direct link to Sameer .K. Banthia's post “What if x goes -1......-2...”
What if x goes -1......-2......-3.....-4 and so on?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Lin Gh.
  Posted 5 years ago. Direct link to Lin Gh.'s post “When you say a function i...”
  When you say a function increases, then you're assuming x is increasing unless otherwise specified.
  Button navigates to signup page
  (5 votes)
Danny Li
Posted 2 years ago. Direct link to Danny Li's post “I need all of you to solv...”
I need all of you to solve this: 2 - X = X^2 + 4
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “2 - x = x^2 + 4 x^2 + 4 +...”
  2 - x = x^2 + 4
  x^2 + 4 + x - 2 = 0
  x^2 + x + 2 = 0
  (x + 2)(x - 1) = 0
  x = -2 and 1
  Comment on Hecretary Bird's post “2 - x = x^2 + 4 x^2 + 4 +...”
  (8 votes)
Ahmed
Posted a year ago. Direct link to Ahmed 's post “Hey Sal how are you? Why ...”
Hey Sal how are you? Why is A going upwards when it should be a negative slop since b is between 0 and 1? Or does the direction of the graph doesn’t effect whether it’s negative or positive?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
Hasan Misra
Posted 8 months ago. Direct link to Hasan Misra's post “Coconuts are awesome !!”
Coconuts are awesome !!
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer

Video transcript

- [Instructor] We're told the function f is defined by the equation above, right over here. Which of the following is the graph of y is equal to negative f of x in the x, y plane? Pause this video and see if you can work through that. All right, now let's work through this together. So first of all, let's think about what y is equal to negative f of x is equal to, well that's going to be the negative of the right-hand side over here. So this is going to be, we could write it as negative 1/3 to the x. And then if we distribute the negative, it'll then be plus one, or we could even write it like this. This is the same thing as one, minus 1/3 to the x. Now there's two ways that you could approach it. You could think about the behavior of this expression as x becomes very large or x becomes very small. Another possibility is we try a table with some easy values. So let me start actually with the table method right over here. So let me just try some easy xs to calculate and figure out what the corresponding y is when y is equal to negative f of x. So one easy one would be, well, we could try zero, but we could see every one of these graphs have the point zero, zero so that won't really help us differentiate between the graphs that well. But let's try the point one and let's try the point negative one. And those are useful because they're pretty easy to calculate. So let's see, when x is equal to one, 1/3 to the first power is just 1/3. One minus 1/3 is 2/3. Now when x is equal to negative one, what's 1/3 to the negative one power? Well, that is equal to three. And so one minus three is equal to negative, negative two. Now let's see, just by inspection, if we can tell which of these graphs contain these two coordinates. So one, 2/3. So one, 2/3. That looks about right. So this one is looking pretty good. B, one, 2/3. This is definitely larger than 2/3, so I would rule that one out. One, 2/3, this looks like negative 2/3. So I'd rule C out. One, 2/3. Definitely nothing right over here, so I'd to rule D out. So A is already looking pretty good, but let's just confirm negative one, negative two. Negative one looks like the point. Negative two is there as well. So I'm feeling very good about A. And you could even check negative one, negative two, the graph doesn't go through there. Negative one, negative two. Doesn't go through there. Negative one, negative two does not go through there. So we really like choice A. Another way that you could have thought about it is when x is very negative, when x is very negative, 1/3 to a very negative power is the same thing as three to a very large power. And so you'll have one minus three to a very large power. Well, three to a very large power is going to be a large value. So one minus, that's gonna be a very negative value. So when x is very negative, your graph should get very negative. And this one checks that box. As x gets more and more negative, this graph is just really getting more and more negative. And then as x gets larger and larger and larger, more and more positive. Well, 1/3 to larger, larger positive exponents is just going to get closer and closer and closer to zero. So you're going to approach one as x gets larger and larger. And it looks like that is what is happening with this graph right over here. As x gets larger and larger and larger, y is approaching one. So once again, we like choice A.