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.