+32 How do you find the range and intercepts of the function f(x)=4|x+5| ?
+2446 The range is the set of all possible outputs that a function can produce.
\(f(x)=4|x+5|\)
The absolute value of any number will can result in a positive number or 0, and 4 multiplied by a positive number or 0 does not change this property at all. Therefore, the range is the following:
\(\text{Range}:\{\mathbb{R}|\hspace{1mm}y\geq0\}\)
This means that the range can be all nonnegative numbers. In interval notation, it would look like the following:
\(\text{Range}:[0,+\infty)\)
How do we find the intercepts? To do it without a graph, you can figure out by setting x=0 and y=0 and solving in each case. Let's do that.
| \(y=4|x+5|\) | Plug in 0 for x. This time, we are solving for the y-intercept. |
| \(y=4|0+5|\) | Simplify inside the absolute value first. |
| \(y=4|5|\) | |
| \(y=4*5=20\) | We have now determined the coordinates of the y-intercept. |
| \((0,20)\) | This is the exact coordinates of the y-intercept. |
Let's do the exact same process. This time, however, we set y=0 to find the x-intecept.
| \(y=4|x+5|\) | Set y equal to 0. |
| \(0=4|x+5|\) | Divide by 4 on both sides. |
| \(0=|x+5|\) | The absolute value always splits an equation into its plus or minus. However, 0 is neither positive nor negative, so there aren't 2 equations that one can set up. |
| \(x+5=0\) | |
| \(x=-5\) | We have now determined the x-intercept, as well. |
| \((-5,0)\) | |
+2446 The range is the set of all possible outputs that a function can produce.
\(f(x)=4|x+5|\)
The absolute value of any number will can result in a positive number or 0, and 4 multiplied by a positive number or 0 does not change this property at all. Therefore, the range is the following:
\(\text{Range}:\{\mathbb{R}|\hspace{1mm}y\geq0\}\)
This means that the range can be all nonnegative numbers. In interval notation, it would look like the following:
\(\text{Range}:[0,+\infty)\)
How do we find the intercepts? To do it without a graph, you can figure out by setting x=0 and y=0 and solving in each case. Let's do that.
| \(y=4|x+5|\) | Plug in 0 for x. This time, we are solving for the y-intercept. |
| \(y=4|0+5|\) | Simplify inside the absolute value first. |
| \(y=4|5|\) | |
| \(y=4*5=20\) | We have now determined the coordinates of the y-intercept. |
| \((0,20)\) | This is the exact coordinates of the y-intercept. |
Let's do the exact same process. This time, however, we set y=0 to find the x-intecept.
| \(y=4|x+5|\) | Set y equal to 0. |
| \(0=4|x+5|\) | Divide by 4 on both sides. |
| \(0=|x+5|\) | The absolute value always splits an equation into its plus or minus. However, 0 is neither positive nor negative, so there aren't 2 equations that one can set up. |
| \(x+5=0\) | |
| \(x=-5\) | We have now determined the x-intercept, as well. |
| \((-5,0)\) | |
