Find the equation of the tangent line to the graph of
f(x) =
2x − 6/x + 1
at the point at which
x = 0.
(Let x be the independent variable and y be the dependent variable.)
I don't know how to find the derivative, i Keep getting it wrong. But I know where to go after that.
Assuming you meant
f = (2x-6)/(x+1)
using the quotient rule,
f' = [(2)(x+1)-(2x-6)(1)]/(x+1)^2
so, at x=0, f' = (2+6)/2 = 4
Or, you can note that
f = (2x-6)/(x+1)
= 2 - 8/(x+1)
so,
f' = 8/(x+1)^2
f'(0) = 8/2 = 4
To find the equation of the tangent line to the graph of a function at a given point, we need to find the slope of the tangent line. To find the slope, we need to find the derivative of the function.
Let's start by finding the derivative of f(x). The function f(x) = (2x - 6) / (x + 1) can be rewritten as f(x) = 2x - 6 * (x + 1)^(-1).
To find the derivative, we can use the quotient rule. The quotient rule states that for a function u(x) / v(x), the derivative is given by (u'(x)v(x) - u(x)v'(x)) / (v(x))^2.
For our function f(x) = 2x - 6 * (x + 1)^(-1), we can let u(x) = 2x - 6 and v(x) = (x + 1)^(-1).
The derivative of u(x) = 2x - 6 is u'(x) = 2, and the derivative of v(x) = (x + 1)^(-1) is v'(x) = -1 * (x + 1)^(-2) * 1.
Applying the quotient rule, we have:
f'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2
= (2 * (x + 1)^(-1) - (2x - 6) * (-(x + 1)^(-2))) / ((x + 1)^(-1))^2
= (2 / (x + 1) + (2x - 6) / (x + 1)^2) / ((x + 1) / 1)^2
= (2 + (2x - 6) / (x + 1)) / (x + 1)^2
Simplifying further, we have:
f'(x) = (2(x + 1) + 2x - 6) / (x + 1)^3
= (4x - 4) / (x + 1)^3
Now that we have the derivative, we can find the slope at the point x = 0.
Substituting x = 0 into the derivative equation, we have:
m = f'(0) = (4(0) - 4) / (0 + 1)^3
= -4 / 1
= -4
So, the slope of the tangent line at the point where x = 0 is -4.
We can now use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Using the point (0, f(0)) = (0, -6) and the slope m = -4, we have:
y - (-6) = -4(x - 0)
y + 6 = -4x
y = -4x - 6
Therefore, the equation of the tangent line to the graph of f(x) at the point where x = 0 is y = -4x - 6.
To find the equation of the tangent line to the graph of a function at a given point, you need to follow these steps:
1. Start by finding the derivative of the function. The derivative represents the slope of the tangent line at any given point on the graph of the function.
In this case, the given function is f(x) = (2x - 6) / (x + 1). To find the derivative, you can use the quotient rule.
The quotient rule states that if you have a function f(x) = g(x) / h(x), then the derivative of f(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Applying the quotient rule to f(x) = (2x - 6) / (x + 1), we get:
f'(x) = [(2 * (x + 1) - (2x - 6) * 1) / (x + 1)^2]
2. Simplify the derivative expression. Expand and simplify the numerator and denominator:
f'(x) = (2x + 2 - 2x + 6) / (x^2 + 2x + 1)
= (8) / (x^2 + 2x + 1)
= 8 / (x^2 + 2x + 1)
3. Plug in the given x-value to find the slope of the tangent line at that point. In this case, x = 0:
f'(0) = 8 / (0^2 + 2(0) + 1)
= 8 / 1
= 8
The slope of the tangent line at x = 0 is 8.
4. Use the point-slope form of a line to find the equation of the tangent line. Since you have the slope (m = 8) and a point (0, f(0)) on the graph, you can use the equation:
y - y1 = m(x - x1)
Plugging in the values, the equation becomes:
y - f(0) = 8(x - 0)
Since x = 0, we have:
y - f(0) = 8(0)
Simplifying further:
y - f(0) = 0
Now, substitute f(0) with the value of the function at x = 0:
y - [f(0)] = 0
y - [f(0)] = 0
Since x = 0, the point on the graph is (0, f(0)).
Hence, the equation of the tangent line is y = f(x) at x = 0.