The figure shows the graph of f(x) = e^x. In Exercises 35-46, use...

Question

The figure shows the graph of f(x) = e^x. In Exercises 35-46, use transformations of this graph to graph each function. Be sure to give equations of the asymptotes. Use the graphs to determine graphs. each function's domain and range. If applicable, use a graphing utility to confirm your hand-drawn  h(x) = e^2x + 1

Accepted Answer

Hello, everyone. In this video, we're going to be looking at this practice problem where we want to graph the equation G of X is equal to E to the seven X plus 12. And we want to graphic using the reference graph of F of X is equal to each of the X. So we want to use transformations to figure out what G of X would look like on the graph. And once we graph it, we want to figure out the ascent towards equations and the domain and range for the function. So in this case, because we're gonna be graphing G of X based off of F of X, let's take a look at the equations and see what the differences are. So F of X is equal to E to the X an exponential function and G of X is equal to E to the seven X plus 12. So you can see that one of the first differences is the seven in front of the X in the exponent. and then we also have the plus 12 after the exponent in G of X. So we want to think about how these two components are affecting the function or how they're changing the function from the standard E of X. So let's think about the seven in the exponent. So the seven in the exponents first off, because there's a coefficient in front of the X on the exponent, we know that it's going to be affecting the horizontal scale. So if we had a coefficient on the outside of the exponents say something like seven times E to the X, this would affect the vertical scale, but we have the coefficient on the exponents. So it's going to affect the horizontal scale. And in this case, because it's a whole number seven, this means it's going to be a compression and a compression bye the inverse of the whole number. So because the whole number seven, the inverse is 1/7 are not the inverse but the reciprocal. So seven is the coefficient in front of X. We do the reciprocal and that's a compression by 1/7. Similarly, if we had had e to the 1/7 X, this is affecting the horizontal still because we have the coefficient in the exponent. But now if we take the reciprocal of 1/7 we have a expansion horizontal expansion By seven because the reciprocal of 1/7 is seven. So now we know how the seven is affecting our equation. Now let's look at this plus 12 that happens after the exponent. So now we can see we're adding a component outside of the exponents. So because it's outside the exponent, I know that it's going to be affecting the Y coordinates or the vertical coordinates. So I'm just gonna go ahead and write vertical here because it's outside B exponent. If it were inside the X one inch, it would look something like E to the seven X plus 12. And I can see that the plus 12 is part of the exponent, but in this case, the plus 12 is on the outside. So it's affecting the vertical scale or the vertical units. And In this case, it's positive 12, which means there's a vertical Translation where we have to go 12 units up. If the 12 was negative, then we would have to go down. But in this case, it's positive. So now we figured out how G of X differs from F of X. So if we use this graph here, that's from option A, I don't know if option is correct, but they give us the F of X function which is E to the X. So I'm gonna build what I think G of X should look like on this first graph. So because there's no negative, there's no sign change between F of X or G of X, there's no negative in front of the G of X. I know that there's going to be no reflections. So I'm expecting to see the same form that has in G of X with a bit of stretches or compressions. So I know I want to see the same you form. But let's try to narrow the search down and already because I'm expecting this same form as F of X. You can see that option A is not going to be the correct answer, but let's continue building this. So we have a compression by 1/7 horizontal compression by 1/7 which means that where we have F of X at this height where Why is equal to 12? Because there's a compression by 1/7, I expect GFX two be closer to the Y axis than a five X. So I'm really expecting G of X to be on the left hand side of F of X. But apart from this horizontal translation, remember that it's going to be 12 units up. So that means that I have to go, you can see the F of X starts on the X axis. Well, G of X needs to start 12 units above that, which would be at y equals 12 and then shoot upwards. So now that I have a better image of what I expect to see, let's look at the answer choices and see if any of those reflect what I'm expecting to see. So A is not, We've already determined is not the one B also has a reflection and C seems to have the same shape But see begins units down And we want 12 units up. So close but not the one that we want. And now we have D and D has the shape that we want. You can see that G of X is on the left Of F of X and it starts 12 units up. So this has the correct shape. So now we can figure out the horizontal ascent oats and the domain and range. So when we think of the horizontal ascent oats, we want to think about where the Y values approach but never pass. And that would be here, You can see that the graph gets really close to y equals 12, but it doesn't pass that marker. So we know that there's a horizontal ascent tote at Y equals 12. And now we want to look at the domain, the domain is the values of X that this function can take. And if we look at this graph, you can see that X could really equal any value, we can double check that by looking at the equation E to the seven X plus 12. And again, we can see on the graph, but we can also confirm it by looking at the equation and there's no value of X that we can plug in that would cause this function to be undefined because even if we plug in negative two, this would become E to the seven times negative two plus 12. This becomes E to the negative 14 plus 12 and E to the negative 14. Is really just one over E to the 14 plus 12. So you can see that this fraction is still a defined output. So any value could really get plugged into X. So that means that the domain is negative infinity to positive infinity. And when we think of range, we think of the Y values. So from this function, You can see that there's no y values below the Y equals 12 line. So that means that although I can get really close to 12, I won't ever equal 12. So that means that I'll have a parentheses and then a lower limit at 12. And then from there, there is no upper limit for Y value. So I can go all the way up to positive infinity. So this becomes my final answer for graphing G of X. And we can see that answer choice D has the shape that we expect. And it also lists the domain as negative infinity to positive infinity, the range as 12 to positive infinity And the horizontal a cento as y equals 12. So hopefully this video helped and see you next time.

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