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SAT (Fall 2023)
Course: SAT (Fall 2023) > Unit 10
Lesson 2: Passport to advanced mathematics- Solving quadratic equations — Basic example
- Solving quadratic equations — Harder example
- Interpreting nonlinear expressions — Basic example
- Interpreting nonlinear expressions — Harder example
- Quadratic and exponential word problems — Basic example
- Quadratic and exponential word problems — Harder example
- Manipulating quadratic and exponential expressions — Basic example
- Manipulating quadratic and exponential expressions — Harder example
- Radicals and rational exponents — Basic example
- Radicals and rational exponents — Harder example
- Radical and rational equations — Basic example
- Radical and rational equations — Harder example
- Operations with rational expressions — Basic example
- Operations with rational expressions — Harder example
- Operations with polynomials — Basic example
- Operations with polynomials — Harder example
- Polynomial factors and graphs — Basic example
- Polynomial factors and graphs — Harder example
- Nonlinear equation graphs — Basic example
- Nonlinear equation graphs — Harder example
- Linear and quadratic systems — Basic example
- Linear and quadratic systems — Harder example
- Structure in expressions — Basic example
- Structure in expressions — Harder example
- Isolating quantities — Basic example
- Isolating quantities — Harder example
- Function notation — Basic example
- Function notation — Harder example
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Nonlinear equation graphs — Harder example
Watch Sal work through a harder Nonlinear equation graphs problem.
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Why are the harder examples always easier than the practice problems?(19 votes)
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(7 votes)
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?(12 votes)
Unfortunately, I don't think so. The problem is that thexis 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.(11 votes)
- Is there another way to find out the answer other than plugging in random numbers?(3 votes)
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.(11 votes)
the harder example is easier than the basic one(6 votes)
What if x goes -1......-2......-3.....-4 and so on?(4 votes)- When you say a function increases, then you're assuming x is increasing unless otherwise specified.(5 votes)
- lets go October sat here baby!! Need a thick 800 math score(4 votes)
- I need all of you to solve this: 2 - X = X^2 + 4(1 vote)
- 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(6 votes)
- I didn’t get why he took -f(x) can some please explain. Can’t we just do it f(x) without the minus(2 votes)
The question is asking to find the graph of -f(x).(4 votes)
- 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?(3 votes)
Shouldn't we mention negative numbers and zero? I know the results are the same but I think it's better if we discuss them too(2 votes)
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.