find the horizontal tangents of the curve and show how.

y=x^3-2x^2+x+1

y = x^3 - 2x^2 + x + 1

Horizontal tangents have slope = 0

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

so, at x = 1/3 or 1 the slope is zero.

So, the lines are

y = 13/9

y = 1

Well, well, well, let's hunt down those horizontal tangents, shall we? To find the horizontal tangents of a curve, we need to locate the points where the derivative equals zero. Let's get our detective hats on and start investigating!

First, we need to find the derivative of the given curve. The derivative of y = x^3 - 2x^2 + x + 1 with respect to x is:

dy/dx = 3x^2 - 4x + 1

Now, we want to find the values of x where the derivative equals zero. So, we set dy/dx = 0 and solve for x:

3x^2 - 4x + 1 = 0

Now, we can solve this quadratic equation to find the values of x. But wait, this looks like a job for the quadratic formula!

x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 3, b = -4, and c = 1. Plug these values into the formula, crunch some numbers, and you'll find the solutions for x.

Once you have those x-values, substitute them back into the original equation y = x^3 - 2x^2 + x + 1 to find the corresponding y-values. And voila! You've got the points where the curve has horizontal tangents.

Remember, while math may seem drab sometimes, adding a little detective work and humor makes it a lot more fun! Happy hunting!

To find the horizontal tangents of the curve y = x^3 - 2x^2 + x + 1, we need to find the points where the derivative of y with respect to x equals zero.

Step 1: Calculate the derivative of y with respect to x:
dy/dx = 3x^2 - 4x + 1

Step 2: Set the derivative equal to zero:
3x^2 - 4x + 1 = 0

Step 3: Solve the quadratic equation for x. We can use the quadratic formula:
x = (-(-4) ± √((-4)^2 - 4(3)(1))) / (2(3))
x = (4 ± √(16 - 12)) / 6
x = (4 ± √4) / 6
x = (4 ± 2) / 6

Step 4: Simplify the solutions:
x1 = (4 + 2) / 6 = 6 / 6 = 1
x2 = (4 - 2) / 6 = 2 / 6 = 1/3

Step 5: Substitute the values of x into the original equation y = x^3 - 2x^2 + x + 1 to find the corresponding y-values:
For x = 1: y1 = 1^3 - 2(1)^2 + 1 + 1 = 1 - 2 + 1 + 1 = 1
For x = 1/3: y2 = (1/3)^3 - 2(1/3)^2 + 1/3 + 1 = 1/27 - 2/9 + 1/3 + 1 = 0

Therefore, the points where the curve has horizontal tangents are (1, 1) and (1/3, 0).

To find the horizontal tangents of a curve, we need to find the points on the curve where the derivative is equal to zero. The derivative represents the rate of change of the function, and at points where the derivative is zero, the function is not increasing or decreasing.

Let's find the derivative of the function f(x) = x^3 - 2x^2 + x + 1 using the power rule:
f'(x) = 3x^2 - 4x + 1.

Now, we need to find the x-values where the derivative is equal to zero. So, at these points, f'(x) = 0:
3x^2 - 4x + 1 = 0.

To solve this quadratic equation, we can factorize or use the quadratic formula. In this case, factoring is a bit challenging, so let's use the quadratic formula to find the x-values:
x = (-b ± √(b^2 - 4ac)) / (2a).

In our equation, a = 3, b = -4, and c = 1. Plugging in these values, we get:
x = (-(-4) ± √((-4)^2 - 4(3)(1))) / (2(3)).
x = (4 ± √(16 - 12)) / 6.
x = (4 ± √4) / 6.

Simplifying further, we have two x-values:
x = (4 ± 2) / 6.
x = 6/6 = 1 (when taking the positive root) or x = 2/6 = 1/3 (when taking the negative root).

So, the x-values for the horizontal tangents of the curve are x = 1 and x = 1/3. To find the corresponding y-values, we substitute these x-values back into the original function:

For x = 1: y = (1)^3 - 2(1)^2 + (1) + 1 = 1 - 2 + 1 + 1 = 1.
For x = 1/3: y = (1/3)^3 - 2(1/3)^2 + (1/3) + 1 ≈ -0.1852.

Therefore, the points (1, 1) and (1/3, -0.1852) represent the two horizontal tangents of the curve y = x^3 - 2x^2 + x + 1.