Logarithmic Functions Changing the Graph

Logarithmic Functions Changing the Graph - Video

Logarithmic Function Model

Logarithmic Function Model:

Here we have what happens when we change the different parts of an exponential function.

When a changes, then there is a vertical shrink/stretch. Shrink occurs when 0 < a < 1 and stretch occurs when a > 1.

When b changes, then the base changes. The overall look does not change just how fast the graph rises.

When c changes, then there is a horizontal shrink/stretch. Shrink occurs when a > 1 and stretch occurs when 0 < a < 1.

When d changes, then there is a horizontal shift (left/right).

When f changes, then there is a vertical shift (up/down).

There is not an e because e is on the most common bases in exponential functions.

Guidelines on How to Graph

Guidelines on How to Graph:

On the left we have 6 different facts on the standard of a logarithmic function and on the right we have the guidelines.

Example 1 - Vertical Stretch/Shrink

Example 1 - Vertical Shrink/Stretch:

Here we have three functions: y = 2 * log(x), y = 1 * log(x), and y = (1/2) * log(x).

We have created an XY table for each one. The best bet is to pick three points: one close to 0, one x-intercept, and one close to the end of the x-axis on your graph. Both the numbers close to 0 and close to the end, if possible, should be multiples of your base, in this case 10, so that you can get integers.

We have (1/10, -2), (1, 0), and (10, 2) for y = 2 * log(x).

We have (1/10, -1), (1, 0), and (10, 1) for y = 1 * log(x).

We have (1/10, -1/2), (1, 0), and (10, 1/2) for y = (1/2) * log(x).

Let's take a closer look at our points. The y-value of each first point is 3/8, 1/8, and 1/24, but the x-value is the same for each one. Why?

We changed the output by multiplying by 2 in the first equation,y = 2 * log(x). If we work backwards, -2 divided by 2 is -1, which is the output value of the second equation, y = 1 * log(x). So the same y-value in the first equation is 2 times higher than the first equation.

We change the output by multiplying by 1/2 in the third equation, y = (1/2) * log(x). If we work backwards, -1/2 times 2 is -1, which is the output value of the second equation, y = (1/2) * log(x). So the same y-value in the third equation is 2 times lower than the second equation.

Now let's take a look at the x-intercept. From the facts, each one has an intercept at (1, 0). The log of 1 rule states that log 1 is equal to 0, anything times 0 is 0. So the a value can be anything in this scenario and the x-intercepts will always be (1, 0).

However, we don't have any y-intercepts, because the standard equation has an asymptote at x = 0. So the graph will not cross the y-axis in this scenario.

As for our domain, we do have an restrictions or limits on our function, because the input values of logarithms can't be negative so the domain of each is (0, +∞).

As for our range, we don't an restrictions or limits on our logarithmic functions, so the range of each function is (-∞, +∞).

Example 2 - Horizontal Stretch/Shrink

Example 2 Horizontal Stretch/Shrink:

Here we have three functions: y = log(2 * x), y = log(1 * x), and y = log([1/2] * x).

We have created an XY table for each one. The best bet is to pick three points: one close to 0, one x-intercept, and one close to the end of the x-axis on your graph. Both the numbers close to 0 and close to the end, if possible, should be multiples of your base, in this case 10, so that you can get integers.

We have (1/20, -1), (1/2, 0), and (5, 1) for y = log(2 * x).

We have (1/10, -1), (1, 0), and (10, 1) for y = log(1 * x).

We have (1/5, -1), (2, 0), and (20, 1) for y = log([1/2] * x).

Let's take a closer look at our points. The x-value of each first point is 1/20, 1/10, and 1/5, but the y-value is the same for each one. Why?

We changed the input by multiplying by 2 in the first equation, y = log(2 * x). If we work backwards, 1/20 times by 2 is 1/10, which is the input value of the second equation, y = log(1 * x). So the same y-value in the third equation is reached 2 times faster than the second equation.

We change the input by multiplying by 1/2 in the third equation, y = log([1/2] * x). If we work backwards, 1/5 times 1/2 is 1/10, which is the input value of the second equation, y = log(1 * x). So the same y-value in the third equation is reached 2 times slower than the second equation.

Now let's take a look at the x-intercept. From the facts, each one has an intercept at (1, 0). The log of 1 rule states that log 1 is equal to 0, but we changed the input values. So the x-intercept of the first equation, y = log(2 * x), is ([1/2], 0). The x-intercept of the second equation, y = log(1 * x), remains the same, (1, 0). The x-intercerpt of third equation, y = log([1/2] * x), is (2, 0).

However, we don't have any y-intercepts, because the standard equation has an asymptote at x = 0. So the graph will not cross the y-axis in this scenario.

As for our domain, we do have an restrictions or limits on our function, because the input values of logarithms can't be negative so the domain of each is (0, +∞).

As for our range, we don't an restrictions or limits on our logarithmic functions, so the range of each function is (-∞, +∞).

Example 3 - Horizontal Shift (Left/Right)

Example 3 Horizontal Shift (Left/Right):

Here we have three functions: y = log(x + 1), y = log(x + 0), and y = log(x - 1).

We have created an XY table for each one. The best bet is to pick three points: one close to 0, one x-intercept, and one close to the end of the x-axis on your graph. Both the numbers close to 0 and close to the end, if possible, should be multiples of your base, in this case 10, so that you can get integers.

We have (-9/10, -1), (0, 0), and (9, 1) for y = log(x + 1).

We have (1/10, -1), (1, 0), and (10, 1) for y = log(x + 0).

We have (11/10, -1), (2, 0), and (11, 1) for y = log(x - 1).

Let's take a closer look at our points. The x-value of each first point is -9/10, 1/10, and 11/10, but the y-value is the same for each one. Why?

We changed the input by moving 1 to the left in the first equation, y = log(x + 1). If we work backwards, -9/10 plus 1 is 1/10, which is the input value of the second equation, y = log(x + 0). So the same x-value in the first equation is reaches the same output 1 number faster than the second equation.

We change the input by moving 1 to the right 1 in the third equation, y = log(x - 1). If we work backwards, 11/10 minus 1 is 1/10, which is the input value of the second equation, y = log(x + 0). So the same x-value in the third equation is reaches the same output 1 number slower than the second equation.

Now let's take a look at the x-intercept. From the facts, each one has an intercept at (1, 0). Our first equation, y = log(x + 1), shifted to the left one, which means the x-intercept moves 1 to the left as well, (0, 0). Our second equation, y = log(x + 0), did not shift, so the x-intercept stays at the same place, (1, 0). Our third equation, y = log(x - 1), shifted to the right one, which means the x-intercept moves 1 to the right as well, (2, 0).

Since we left shifted the first equation, y = log(x + 1), to the left, we now have a y-intercept at (0, 0) because our asymptote as shifted as well. The second and third equation still do not cross the y-axis.

As for our domain, we do have an restrictions or limits on our function. The first equation, y = log(x + 1), shifted to the left 1 so our domain shifted to (-1, +∞). The second equation, y = log(x + 0), did not shift so our domain remains the same, (0, +∞). And, our third equation, y = log(x - 1), shifted to the right 1 so our domain shifted to (1, +∞).

As for our range, we don't an restrictions or limits on our logarithmic functions, so the range of each function is (-∞, +∞).

Example 4 - Vertical Shift (Up/Down)

Example 4 Vertical Shift (Up/Down):

Here we have three functions: y = log(x) + 2, y = log(x) + 0, and y = log(x) - 2.

We have created an XY table for each one. The best bet is to pick three points: one close to 0, one x-intercept, and one close to the end of the x-axis on your graph. Both the numbers close to 0 and close to the end, if possible, should be multiples of your base, in this case 10, so that you can get integers.

We have (1/10, 1), (1, 2), and (10, 3) for y = log(x) + 2.

We have (1/10, -1), (1, 0), and (10, 1) for y = log(x) + 0.

We have (1/10, -3), (1, -2), and (10, -1) for y = log(x) - 2.

Let's take a closer look at our points. The y-value of each first point is 1, -1, and -3, but the x-value is the same for each one. Why?

We changed the output by adding 2 in the first equation, y = log(x) + 2. If we work backwards, 1 minus 2 is -1, which is the output value of the second equation, y = log(x) + 0. So the same y-value in the first equation is 2 more than the second equation.

We change the output by subtracting 2 in the third equation, y = log(x) - 1. If we work backwards, -3 plus 2 is -1, which is the output value of the second equation, y = log(x) + 0. So the same y-value in the third equation is 2 fewer than the second equation.

Now let's take a look at the x-intercept. From the facts, each one has an intercept at (1, 0). Our first equation, y = log(x) + 2, shifted up 2, which means the x-intercept has changed to (1/100, 0). We can rewrite 1/100 as 1/10^-2. Notice the 2. Our second equation, y = log(x) + 0, did not shift, which means the x-intercept has remained the same (1, 0). Our third equation, y = log(x) - 2, shifted down 2, which means the x-intercept has changed to (100, 0). We can rewrite 100 as 10². Notice the 2.

However, we don't have any y-intercepts, because the standard equation has an asymptote at x = 0. So the graph will not cross the y-axis in this scenario.

As for our domain, we do have an restrictions or limits on our function, because the input values of logarithms can't be negative so the domain of each is (0, +∞).

As for our range, we don't an restrictions or limits on our logarithmic functions, so the range of each function is (-∞, +∞).

Example 5 - Reflection Across X-axis

Example 5 Reflection Across X-axis:

Here we have three functions: y = 1 * log(x), y = -1 * log(x).

We have created an XY table for each one. The best bet is to pick three points: one close to 0, one x-intercept, and one close to the end of the x-axis on your graph. Both the numbers close to 0 and close to the end, if possible, should be multiples of your base, in this case 10, so that you can get integers.

We have (1/10, -1), (1, 0), and (10, 1) for y = 1 * log(x).

We have (1/10, 1), (1, 0), and (10, -1) for y = -1 * log(x).

Let's take a closer look at our points. The y-value of each first point is 1 and -1 but the x-value is the same for each one. Why?

We changed the output by multiplying by -1 in the second equation, y = -1 * log(x). If we work backwards, 1 times -1 is -1, which is the output value of the first equation, y = 1 * log(x). So the same y-value in the second equation is the negative of the first equation.

Now let's take a look at the x-intercept. The first equation, y = 1 * log(x), did not shift and is a standard logarithmic equation so the asymptote stayed the same at x = 0 so our x-intercept is (1, 0). For the second equation, y = -1 * log(x), is also (1, 0).

We do not have a y-intercept.

As for our domain, the first equation is standard and did not move so the domain of each is (0, +∞). Our second equation flipped over the x-axis, so the inputs remained the same. So, our domain is still (-∞, 0).

As for our range, we don't an restrictions or limits on our logarithmic functions, so the range of each function is (-∞, +∞).

Example 6 - Reflection Across Y-axis

Example 6 Reflection Across Y-axis:

Here we have three functions: y = log(1 * x), y =log([-1] * x).

We have created an XY table for each one. The best bet is to pick three points: one close to 0, one x-intercept, and one close to the end of the x-axis on your graph. Both the numbers close to 0 and close to the end, if possible, should be multiples of your base, in this case 10, so that you can get integers.

We have (1/10, -1), (1, 0), and (10, 1) for y = log(1 * x).

We have (-1/10, -1), (-1, 0), and (-10, 1) for y = log([-1] * x).

Let's take a closer look at our points. The x-value of each first point is 1/10 and -1/10, but the y-value is the same for each one. Why?

We changed the input by multiplying by -1 in the second equation, y = log([-1] * x). If we work backwards, -1/10 divided by -1 becomes 1/10. 1/10 is the output of the first equation, y = log(1 * x). So the same y-value in the first equation is the reciprocal of the second equation.

Now let's take a look at the x-intercept. The first equation, y = log(1 * x), did not shift and is a standard logarithmic equation so the asymptote stayed the same at x = 0 so our x-intercept is (1, 0). For the second equation, y = -1 * log(x), has changed to (-1, 0).

We do not have a y-intercept.

As for our domain, the first equation is standard and did not move so the domain of each is (0, +∞). Our second equation flipped over the y-axis, so the inputs are now negative numbers. So, our domain is (-∞, 0).

As for our range, we don't an restrictions or limits on our logarithmic functions, so the range of each function is (-∞, +∞).

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