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# Make a rough sketch of the graph of each function. Do not use a calculator. Just use the graphs given in Figures 12 and 13 and, if necessary, the transformations of Section 1.3.(a) $y = \log_{10} (x + 5)$(b) $y = -\ln x$

## (a) Shift the graph of $y=\log _{10} x$ five units to the left toobtain the graph of $y=\log _{10}(x+5) .$ Note the verticalasymptote of $x=-5$.(b) Reflect the graph of $y=\ln x$ about the $x$ -axis to obtainthe graph of $y=-\ln x$.

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### Video Transcript

we're going to make a rough sketch of this graph. So let's start by thinking about the standard function. Why equals log based 10 of X. Just a rough sketch of that. We know the lager with McGrath shape for longer than the growth we know this one would go through the 10.10 and also down the line. Somewhere down there, it would go through the 0.10 1 So what effect does it have if we add five to the X, that transformation is going to shift. It left five. So let's take the graph. We just drew and shipped. It left five. Now, remember what we just drew has a vertical Assen tote the line X equal zero. So if we should that to the left five. We now have a vertical Lassen toed the line X equals negative five. And if we take our X intercept, which was 10 and we shifted to the left five, it's now over here at negative 40 And if we take our 0.10 1 and shifted to the left five, we get the point 51 So there's a rough sketch of the graph, and now for Part B. The same idea. Let's start with a parent function. Why equals natural log of X? We know it's basic shape, Same basic shape as the previous graph also has an X intercept at 10 this one goes through the point E one. Now what effect does the negative have on it that's going to reflect it across the X axis? All right, so let's draw that. So the point at 10 is still going to be at 10 We're just flipping the whole thing upside down. The point that was previously E one is now e negative one.

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