# My question: An equation of the line tangent to the graph of f(x) = x(1 - 2x)3 at the point (1, –1) is??

My answer: y = –2x + 1

## take the derivative using the product rule

f ' (x) = x(3)(1-2x)^2 (-2) + (1-2x)^3

sub in the x of the point(1,-1)

f ' (1) = 1(3)(1)(-2) + (-1)

= -7 which will be the slope of your tangent

Now you have the slope of -7 and a point (1,-1)

Use your favourite method of finding the equation of the line.

## Final answer would be y=-7x+6 one way to figure it out is by finding you slope which is -7. You will then want to use the equation y=mx+b. We have the coordinates, (1,-1). The slope, -7 is m. 1 is x and -1 is y. Plug it into the equation which is -1=-7(1)+b. Now solve for b. -7(1)=-7

-1=-7+b. Plus 7 on both sides. -1+7= 6.

Y=mx+b, Y=-7x+6 =)

## Okay, thanks

## 我日您爹

## Why did the tangent cross the road?

Because it wanted to be close to the graph of f(x) = x(1 - 2x)3!

## To find the equation of the line tangent to the graph of the function f(x) = x(1 - 2x)^3 at the point (1, -1), we need to find the slope of the tangent line at that point and then use the point-slope form of a linear equation.

Step 1: Find the derivative of the function f(x) with respect to x. The derivative represents the slope of the tangent line at any given point on the graph.

Taking the derivative of f(x) = x(1 - 2x)^3 involves using the product and chain rules. The derivative is given by:

f'(x) = [1*(1 - 2x)^3] + [3*x*(-2)*(1 - 2x)^2]

= (1 - 2x)^2 - 6x(1 - 2x)^2

= (1 - 2x)^2[1 - 6x]

Step 2: Plug in x = 1 into the derivative to find the slope of the tangent line at the point (1, -1).

f'(1) = (1 - 2(1))^2[1 - 6(1)]

= (1 - 2)^2[1 - 6]

= (-1)^2[-5]

= 1(5)

= 5

So, the slope of the tangent line at (1, -1) is 5.

Step 3: Use the point-slope form of a linear equation to find the equation of the tangent line.

The point-slope form of a linear equation is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) are the coordinates of a point on the line.

Using the point (1, -1) and the slope 5, we have:

y - (-1) = 5(x - 1)

y + 1 = 5x - 5

y = 5x - 6

So, the equation of the line tangent to the graph of f(x) = x(1 - 2x)^3 at the point (1, -1) is y = 5x - 6.