Sketch the curve with equation

y = (x -1)³ - 4(x - 1)

It's not really the sketching which is the problem but more working out where the graph crosses the x-axis and y-axis? I can see that it crosses the x-axis at 1 but am clueless as to the other two parts at which it crosses the x-axis .. thanks!

well, try factoring it

y = (x-1)((x-1)^2 - 4)
= (x-1)(x-1+2)(x-1-2)
= (x-1)(x+1)(x-3)

That help?

Naturally, the y-intercept (at x=0) is at y=3

To find out where the graph of the equation y = (x - 1)³ - 4(x - 1) crosses the x-axis, we need to determine the x-values where the function's output, y, equals 0.

To solve for x, we set y equal to 0 and solve the equation:

0 = (x - 1)³ - 4(x - 1)

Expanding the brackets, we get:

0 = (x - 1)(x - 1)(x - 1) - 4(x - 1)

We can simplify this expression:

0 = (x - 1)³ - 4(x - 1)

Now, we can factor out the common factor (x - 1) from both terms:

0 = (x - 1)[(x - 1)² - 4]

Next, we can further simplify the expression inside the square bracket:

0 = (x - 1)[(x - 1)(x - 1) - 4]

0 = (x - 1)[(x² - 2x + 1) - 4]

0 = (x - 1)[x² - 2x - 3]

Now we have two factors: (x - 1) and (x² - 2x - 3).

To find the x-values where the graph crosses the x-axis, we equate each factor to zero and solve for x:

First factor: x - 1 = 0
Solving this equation, we find that x = 1.

Now let's solve for the second factor:

x² - 2x - 3 = 0

Since this equation cannot be factored easily, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = -2, and c = -3.

Plugging these values into the quadratic formula, we get:

x = (2 ± √((-2)² - 4(1)(-3))) / (2(1))

Simplifying further:

x = (2 ± √(4 + 12)) / 2

x = (2 ± √16) / 2

x = (2 ± 4) / 2

For the "+" case:

x = (2 + 4) / 2

x = 6 / 2

x = 3

For the "-" case:

x = (2 - 4) / 2

x = -2 / 2

x = -1

Therefore, the graph of the equation y = (x - 1)³ - 4(x - 1) crosses the x-axis at three points: x = 1, x = 3, and x = -1.

To sketch the curve, you can plot these x-values on a coordinate plane and substitute them back into the original equation to find the corresponding y-values. After plotting these points, you can connect them to form the curve.