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The graphs of the linear function f of x is equal to mx plus b, and the exponential function g of x is equal to a times r to the x, where r is greater than 0 pass through the points negative 1 comma 9-- so this is negative 1 comma 9 right over here-- and 1 comma 1. Both graphs are given below. So this very clearly is the linear function, it is a line right over here. And this right over here is the exponential function. So right over here. And given the fact that this exponential function keeps decreasing as x gets larger and larger and larger, is a pretty good hint that our r right over here, they tell us r is greater than 0, but it's a pretty good hint that r is going to be between 0 and 1. The fact that g of x keeps approaching, it's getting closer and closer and closer to 0 as x increases. But let's use the data they're giving us, the two points of intersection to figure out what the equations of these two functions are. So first we can tackle the linear function. So f of x is equal to mx plus b. So they give us two points, we could use those points first to figure out the slope. So our m, right over here, that's our slope, that's our change in y over change in x. The rate of change of the vertical axis with respect to the horizontal axis. So let's see, between those two points, what is our change in x? Our change in x, if we were going from x equals negative 1 to x equals 1. So we could think of it as, we're finishing at 1, we started at negative 1. So 1 minus negative 1, our change in x is 2. We see that right over there. And what about our change in y? Well, we start at 9. Let me do this in maybe another color here just-- We start at 9, and we end up at 1. So we end up at 1. We started at 9. 1 Minus 9 is negative 8. And just to be clear, when x is 1, y is 1. When x is negative 1, y is 9. Another way to think about it, the way I drew it right over here, we're finishing at x equals 1, y equals 1. We started at x equals negative 1, y equals 9. And so we just took the differences, we get negative 8 over 2, which is equal to negative 4. And so now we can write that f of x is equal to negative 4, that's our slope, times x. Negative 4 times x plus b. And you can see that slope right over here. Every time you increase your x by 1-- you got to be careful here-- but every time you increase your x by 1, you're decreasing your y. And here on the x-axis, we're marking off every 1/2. So every time you increase your x by 1, you are decreasing your y by 4 there. So that makes sense that the slope is negative 4. So now let's think about what b is. So to figure out b, we could use either one of these points to figure out, given an x, what f of x is. And then we can solve for b. So let's try f of, let's try 1, because 1 is a nice simple number. So we could write f of 1, which would be negative 4 times 1 plus b. And they tell us that f of 1 is 1, is equal to 1. And so this part right over here, we could write that as negative 4 plus b is equal to 1. And then we could add 4 to both sides of this equation, and then we get b is equal to 5. So we get f of x is equal to negative 4 x plus 5. And now does that make sense, that the y-intercept here is 5? Well, you see that right over here by inspection, you could have guessed actually that the y-intercept here is 5. now we've solved it. Maybe this was 5.00001 or something, but now we know for sure. It's negative 4 x plus 5. Or another way you could have said it, if the slope is negative 4, if this right over here is 9, you increase 1 in the x direction, you're going to decrease 4 in the y direction. That will get you to y is equal to 5. So that is the y-intercept. But either way, we have figured out the linear function. Now let's figure out the exponential function. So here we could just use the two points to figure out these two unknowns. So for example, let's just try this first point. So g of negative 1, which if we look at this right over here, would be a times r to the negative 1. They tell us that g of negative 1 is going to be equal to 9. And so, we could write this a times r to the negative 1, that's the same thing as a over r is equal to 9. Or we can multiply both sides by r. And we could say a is equal to 9r. Now let's use this other point. They tell us that g of 1, which would be the same thing as a times r to the first power, or just a times r, that that is equal to 1. Or a times r is equal to 1. So how can we use this information right here, a is equal to 9r and a times r is equal to 1, to solve for a and r? Well, I have a little system here, it is a non-linear system, but it's a pretty simple one. We could just take this a and substitute it in right over here for a. And so we would get 9r for a. This first constraint tells us a must be equal to 9r. So 9r, instead of writing in a here, I'll write 9r times r is equal to 1. Or we could write-- let me scroll down a little bit-- we could write 9r squared is equal to 1. Divide both sides by 9. r squared is equal to 1 over 9. And now to figure out r, you might want to take the positive and negative square root of both sides. But they tell us that r is greater than 0. So we can just take the principal root of both sides, and we get r is equal to 1/3. And then we could substitute this back into either one of these other two to figure out what a is. We know that a is equal to 9 times r. So 9 times 1/3, a is equal to 3. So our exponential function could be written as g of x is equal to a, which is 3, times r, which is 1/3 to the 1/3, to the x power. Or you could even write it like this because order of operations would imply that you do the exponential first. Watch, I'll write it like this again, just to make it clear this isn't just 1 to the third power, this is 1/3 to the third power.