How do you find the equation of a line tangent to the function y=x^2(x-2)^3 at x=1?

2 Answers
May 12, 2018

The equation is y=9x-10.

Explanation:

To find the equation of a line, you need three pieces: the slope, an x value of a point, and a y value.

The first step is to find the derivative. This will give us important information about the slope of the tangent. We will use the chain rule to find the derivative.

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

The derivative tells us the points what the slope of the original function looks like. We want to know the slope at this particular point, x=1. Therefore, we simply plug this value into the derivative equation.

y=3(1)^2(1-2)^2
y=9(1)
y=9

Now, we have a slope and an x value. To determine the other value, we plug x into the original function and solve for y.

y=1^2(1-2)^3
y=1(-1)
y=-1

Therefore, our slope is 9 and our point is (1,-1). We can use the formula for the equation of a line to get our answer.

y=mx+b

m is the slope and b is the vertical intercept. We can plug in the values we know and solve for the one we don't.

-1=9(1)+b
-1=9+b
-10=b

Finally, we can construct the equation of the tangent.

y=9x-10

I have solved this way! Please, see the answer below:
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