find the tangent to y=27/(x^2+2) at (1,9)
y = 27(x^2+2)^-1
dy/dx = -27(2x)(x^2+2)^-2 = -54x/(x^2+2)^2
at x = 1 , dy/dx = -54/9 = -6
equation:
y = -6x+b , sub in the point (1,9)
9 = -6+b
b = 15
tangent equation: y = -6x + 15
Well, well, well... Are you ready for some tangential comedy? Brace yourself, because here it comes!
Let's find the tangential wit to the curve y = 27/(x^2 + 2) at the point (1,9).
First, we need to find the derivative of this function. Using the power rule, the derivative would be:
y' = -54x/(x^2 + 2)^2
Now, plug in x = 1 to find the slope of the tangent at (1,9):
y'(1) = -54(1)/(1^2 + 2)^2
= -54/9
= -6
So, the slope of our tangential humor is -6.
Now, we can use the point-slope equation to find the equation of the tangent:
y - 9 = -6(x - 1)
Simplifying further, we get:
y = -6x + 15
And there you have it! The tangent to the curve y = 27/(x^2 + 2) at the point (1,9) is y = -6x + 15. Just beware, this tangent might be filled with witty zingers and hilarious punchlines!
To find the tangent line to the curve y = 27/(x^2 + 2) at the point (1, 9), we need to determine the slope of the tangent line at that point and then use the point-slope form of the equation of a line.
Let's start by finding the derivative of the function y with respect to x, denoted as dy/dx.
Taking the derivative of y = 27/(x^2 + 2) using the quotient rule, we get:
dy/dx = [27 * (d/dx)(x^2 + 2) - (x^2 + 2) * (d/dx)(27)] / (x^2 + 2)^2
Simplifying this, we have:
dy/dx = [27 * (2x) - 0] / (x^2 + 2)^2
= 54x / (x^2 + 2)^2
Now, substitute x = 1 into the derivative to find the slope at the point (1, 9):
dy/dx| x=1 = 54(1) / (1^2 + 2)^2
= 54 / 9
= 6
So, the slope of the tangent line at (1, 9) is 6.
Next, we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is:
y - y₁ = m(x - x₁)
Where m is the slope and (x₁, y₁) is the given point.
Plugging in the values, the equation becomes:
y - 9 = 6(x - 1)
Simplifying this expression gives:
y - 9 = 6x - 6
Rewriting in the standard form of a linear equation:
6x - y = 3
So, the tangent line to the curve y = 27/(x^2 + 2) at the point (1, 9) is given by the equation 6x - y = 3.
To find the tangent line to the curve at a given point, we need to find the derivative of the function and then substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line. Finally, we can use the point-slope equation of a line to find the equation of the tangent line.
Step 1: Find the derivative of the function y = 27/(x^2 + 2).
To find the derivative, we can use the quotient rule, where if we have a function in the form f(x) = g(x)/h(x), the derivative is given by f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2.
Let's differentiate each part of the function:
g(x) = 27 and h(x) = x^2 + 2.
g'(x) = 0 (since the derivative of a constant is zero)
h'(x) = 2x.
Using the quotient rule, we have:
y' = [(0 * (x^2 + 2)) - (27 * 2x)] / ((x^2 + 2)^2)
= (-54x) / ((x^2 + 2)^2)
= -54x / (x^4 + 4x^2 + 4).
Step 2: Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line.
The given point is (1, 9).
Substituting x = 1 into the derivative y', we have:
y' = -54(1) / (1^4 + 4(1^2) + 4)
= -54 / 9
= -6.
So, the slope of the tangent line at the point (1, 9) is -6.
Step 3: Use the point-slope equation of a line to find the equation of the tangent line.
The point-slope equation of a line is given by y - y₁ = m(x - x₁), where (x₁, y₁) is the given point on the line and m is the slope.
Using the given point (1, 9) and slope m = -6, we have:
y - 9 = -6(x - 1).
Expanding and simplifying the equation, we get:
y - 9 = -6x + 6
y = -6x + 15.
Therefore, the equation of the tangent line to the curve y = 27/(x^2 + 2) at the point (1, 9) is y = -6x + 15.