Help!!!! I have no clue on how to figure out graphing functions like f(x)=(x-1)^3 - 5

Question

jay_17 · Answer

How do you graph these equations???? Algebra 2

jim_thompson5910 · Answer

If we plugged in x = 0, this is what would happen

$$\Large f(x) = (x-1)^3 - 5$$
$$\Large f(0) = (0-1)^3 - 5$$
$$\Large f(0) = (-1)^3 - 5$$
$$\Large f(0) = -1 - 5$$
$$\Large f(0) = -6$$

So the input of x = 0 leads to the output of -6

This means we have x = 0 and y = -6 pair up together. They form the point (0,-6)

The point (0,-6) is one of the many points on this function curve.

jim_thompson5910 · Answer

(0,-6) is also the y intercept because the function crosses the y axis at this point

jim_thompson5910 · Answer

Now let's plug in x = 1. If we do that, we get

$$\Large f(x) = (x-1)^3 - 5$$
$$\Large f(1) = (1-1)^3 - 5$$
$$\Large f(1) = (0)^3 - 5$$
$$\Large f(1) = 0 - 5$$
$$\Large f(1) = -5$$

So the input of x = 1 leads to the output of -5

This means we have x = 1 and y = -5 pair up together. They form the point (1,-5)

So (1,-5) is another point on this function curve.

jim_thompson5910 · Answer

Repeat this process many times to generate a bunch of points. After you have enough points, plot them on the same grid. Then draw a curve through all the points. The more points you plot, the more accurate the graph.

heyitsjess28 · Answer

You could also think of it this way....

We know that the standard graph of y=x^3 is:
|dw:1472428303001:dw|

If we add/subtract anything within the cubed part of the equation, then the whole graph either moves left/right of the origin. Since we are subtracting 1, the graph moves to the right 1 unit. The graph should look like this:
|dw:1472428615397:dw|

Then, anything added/subtracted outside of the cubed part moves the graph up/down. Since the equation is being subtracted by 5, the whole graph moves down 5 units.
The end result looks like this:
|dw:1472428777032:dw|