Find an equation of the tangent line to the curve at (8,0). y=ln(x^2 -8x +1)
dy/dx = (2x - 8)/(x^2 - 8x + 1)
at (8,0) , dy/dx = (16-8)/(64-64+1) = 8
so for y = mx + b
y = 8x + b
at (8,0),
0 = 64 + b, ---> b = -64
equation: y = 8x - 64
Well, well, well, it seems you've got yourself a curve and you're looking for a tangent line. How exciting! Let's see what we can do here.
To find the equation of the tangent line, we'll need two things: the slope of the tangent line and a point on that line. Lucky for you, we already have the point (8,0) on the curve. Isn't that convenient?
Now, let's talk about the slope. To find the slope of the tangent line, we can take the derivative of the function. So, let's differentiate our function y = ln(x^2 - 8x + 1) with respect to x.
The derivative of ln(x) is 1/x, so applying the chain rule gives us:
dy/dx = (1/(x^2 - 8x + 1))*(2x - 8)
Goodness gracious, that's a mouthful! But fear not, we don't need to simplify it just yet. We're just looking for the slope at this point.
So, now we've got the slope, which is dy/dx. And we have our point (8,0). Oh, it's getting exciting now!
To find the equation of the tangent line, we'll use the point-slope form. The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line.
Plugging in our values, we get:
y - 0 = (1/(8^2 - 8(8) + 1))*(2(8) - 8)(x - 8)
Don't panic, we're almost there! Let's simplify this equation:
y = (1/(65))*(16)(x - 8)
And there you have it! The equation of the tangent line to the curve y = ln(x^2 - 8x + 1) at the point (8,0) is:
y = (16/65)(x - 8)
Ta-dah!
To find the equation of the tangent line to the curve y = ln(x^2 - 8x + 1) at the point (8, 0), we need to find the derivative of the function first.
Step 1: Find the derivative of the function y = ln(x^2 - 8x + 1).
Using the chain rule, we have:
dy/dx = (1 / (x^2 - 8x + 1)) * (2x - 8)
= (2x - 8) / (x^2 - 8x + 1)
Step 2: Substitute the x-coordinate of the given point (8, 0) into the derivative to find the slope of the tangent line.
Let x = 8:
m = (2(8) - 8) / (8^2 - 8(8) + 1)
= (16 - 8) / (64 - 64 + 1)
= 8 / 1
= 8
So, the slope of the tangent line is 8.
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by: y - y1 = m(x - x1).
Taking the point (8, 0) and the slope m = 8:
y - 0 = 8(x - 8)
This equation simplifies to:
y = 8x - 64
Therefore, the equation of the tangent line to the curve y = ln(x^2 - 8x + 1) at the point (8, 0) is y = 8x - 64.
To find the equation of the tangent line to the curve at the point (8, 0), we need to find the slope of the tangent line and use the point-slope form of a line.
Step 1: Find the derivative of the function y = ln(x^2 - 8x + 1).
We can use the chain rule to simplify this process.
Let u = x^2 - 8x + 1.
Then, y = ln(u).
Using the chain rule, dy/du = 1/u * du/dx.
To find du/dx, we differentiate u with respect to x:
du/dx = d/dx(x^2 - 8x + 1)
= 2x - 8.
Substituting this result back into the chain rule, we have:
dy/dx = dy/du * du/dx
= 1/u * (2x - 8).
Step 2: Find the slope of the tangent line at (8, 0).
Plug x = 8 into the derivative we just found:
dy/dx = 1/(8^2 - 8(8) + 1) * (2(8) - 8)
= 1/(64 - 64 + 1) * (16 - 8)
= 1/1 * 8
= 8.
So, the slope of the tangent line at (8, 0) is 8.
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, and m is the slope.
We know that the point (x₁, y₁) is (8, 0), and the slope (m) is 8.
Plugging this information into the formula, we have:
y - 0 = 8(x - 8).
Simplifying further:
y = 8x - 64.
Therefore, the equation of the tangent line to the curve y = ln(x^2 - 8x + 1) at the point (8, 0) is y = 8x - 64.