Find the equation of the tangent line to the graph of

f(x) = (4x − 8)/(x + 1)
at
x = 0.
(Use y for your dependent variable.)

y = (4x-8)/(x+1)

y' = 12/(x+1)^2
So,
y(0) = -8
y'(0) = 12

Now you have a point (0,-8) and a slope (12), so the tangent line is

y = 12x-8

See

http://www.wolframalpha.com/input/?i=plot+y%3D%284x-8%29%2F%28x%2B1%29%2Cy%3D12x-8+for+x+%3D+-1+to+3

Well, it seems like you're looking for the equation of a tangent line to a curve. Let's dive into it!

First, we need to find the derivative of the function f(x). By using the quotient rule, we differentiate the numerator and the denominator separately, and then apply the chain rule:

f'(x) = [(4)(1)(x + 1) - (4x - 8)(1)] / (x + 1)^2

After simplifying, we get:

f'(x) = 12 / (x + 1)^2

Now, to find the equation of the tangent line at x = 0, we need two pieces of information: the slope and a point on the line.

Let's start by finding the slope. We substitute x = 0 into the derivative:

f'(0) = 12 / (0 + 1)^2
f'(0) = 12

So, the slope of the tangent line is 12.

Next, we need a point on the line. We can use the coordinates (0, f(0)):

f(0) = (4(0) - 8) / (0 + 1)
f(0) = -8

Therefore, the point (0, -8) lies on the tangent line.

Using the point-slope form of a line (y - y1 = m(x - x1)), we can plug in our values:

y - (-8) = 12(x - 0)
y + 8 = 12x

So, the equation of the tangent line to the graph of f(x) at x = 0 is:

y = 12x - 8.

Hope that helps! And if you're still confused, don't worry, math can be tangentially confusing at times. Keep solving equations and stay on the right path!

To find the equation of the tangent line to the graph of f(x) at x = 0, we can follow these steps:

Step 1: Find the derivative of f(x) with respect to x.
Step 2: Substitute x = 0 into the derivative to find the slope of the tangent line.
Step 3: Use the point-slope form of the equation of a line to find the equation of the tangent line.

Step 1: Find the derivative of f(x).
To find the derivative, we can use the quotient rule:
f'(x) = [(g(x) * f'(x)) - (f(x) * g'(x))] / [g(x)]^2

In our case, f(x) = (4x - 8)/(x + 1).
Let g(x) = (x + 1).
So, f'(x) = [(g(x) * f'(x)) - (f(x) * g'(x))] / [g(x)]^2
= [(x + 1) * f'(x) - (4x - 8) * 1] / (x + 1)^2

To find f'(x), we differentiate the numerator and denominator separately:
f'(x) = [(1) * (4x - 8) - (4) * (1)] / (x + 1)^2
= (4x - 8 - 4) / (x + 1)^2
= (4x - 12) / (x + 1)^2

Step 2: Substitute x = 0 into the derivative to find the slope of the tangent line.
f'(0) = (4 * 0 - 12) / (0 + 1)^2
= -12 / 1
= -12

Step 3: Use the point-slope form of the equation of a line to find the equation of the tangent line.
The point-slope form of the equation of a line is: y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line.

We know that the point (x1, y1) is (0, f(0)).
Substituting x = 0 into f(x), we get:
y1 = f(0) = (4 * 0 - 8) / (0 + 1)
= -8

So, the point (x1, y1) is (0, -8).

Now, using the slope m = -12 and the point (0, -8), we can write the equation of the tangent line:
y - (-8) = -12(x - 0)
y + 8 = -12x
y = -12x - 8

Therefore, the equation of the tangent line to the graph of f(x) = (4x - 8)/(x + 1) at x = 0 is y = -12x - 8.

To find the equation of the tangent line to the graph of the function f(x) at a specific point, we need to calculate the slope of the tangent line and use the point-slope form of a linear equation.

First, let's find the slope of the tangent line at x = 0. The slope of the tangent line is equal to the derivative of the function evaluated at that point.

To find the derivative of the function f(x), we can use the quotient rule. The quotient rule states that if we have a function of the form h(x) = u(x) / v(x), then its derivative is given by:

h'(x) = (u'(x)v(x) - v'(x)u(x)) / (v(x))^2

For our function f(x) = (4x − 8)/(x + 1), we have u(x) = 4x − 8 and v(x) = x + 1. Let's find u'(x) and v'(x):

u'(x) = 4 (the derivative of 4x − 8 is 4)
v'(x) = 1 (the derivative of x + 1 is 1)

Now, let's apply the quotient rule:

f'(x) = (u'(x)v(x) - v'(x)u(x)) / (v(x))^2
= ((4)(x + 1) - (1)(4x − 8)) / (x + 1)^2
= (4x + 4 - 4x + 8) / (x + 1)^2
= 12 / (x + 1)^2

So, the derivative of f(x) is 12 / (x + 1)^2.

Now, let's evaluate this derivative at x = 0 to find the slope of the tangent line at that point:

f'(0) = 12 / (0 + 1)^2
= 12 / 1
= 12

Therefore, the slope of the tangent line at x = 0 is 12.

Next, we need to find the y-coordinate of the point on the graph corresponding to x = 0. To do this, substitute x = 0 into the original function f(x):

f(0) = (4(0) − 8)/(0 + 1)
= -8/1
= -8

Therefore, the point on the graph corresponding to x = 0 is (0, -8).

Now that we have the slope of the tangent line (m = 12) and a point on the line (0, -8), we can 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 (x1, y1) is a point on the line and m is the slope of the line.

Plugging in the values, we get:

y - (-8) = 12(x - 0)
y + 8 = 12x

Simplifying, we can rewrite this equation in slope-intercept form:

y = 12x - 8

Therefore, the equation of the tangent line to the graph of f(x) = (4x − 8)/(x + 1) at x = 0 is y = 12x - 8.