Question

Find the point where the tangent line to y= x^2 at x=0.8 crosses the x-axis.

x-intercept = ?

Answer 1

Basically. If you can get the equation of the tangent line, then you're just finding the x-intercept of that tangent line. Do you know how to get the tangent line?

Answer 2

The tangent line is y=2x right? o.o

Answer 3

Thats just the derivative.

Answer 4

The derivative will help you find the slope of the tangent line you need, though. So you need that y = 2x.

Once you take the derivative of a function, you pretty much have a formula for the slope at any point along that function. So we want a line tangent to x^2 at x = 0.8. Basically that means we need the slope of x^2 at thepoint 0.8. Well, the derivative gives us a way to find that. Once we have the derivative, 2x, we plug in the x-coordinate, 0.8, and that gives us the slope of the tangent line. So slope will be 1.6.

Now with any line, you need a slope and you need a point so you can make the equation. We have the slope, now we just need the point. Well, it tells us the x-coordinate of the point is 0.8, so if we plug thatinto the function we'll get the y-coordinate

(0.8)^2 = .64.

This means the slope is 1.6 and the point is (0.8, .64). Now this information goes into the point slope form equation, which is

\[y-y _{1}=m(x-x_{1})\] where m is the slope and x1 y1 are coordinates. So plugging in all that info we get

y - .64 = 1.6(x - .8)
y = 1.6x -.64

Now we just find the x-intercept by setting y equal to 0

0 = 1.6x - .64
1.6x = .64
x = .4

Answer 5

Thanks so much I sort of realized the answer, but your response makes it a lot better to understand. Thanks ^^

Answer 6

Np :3