Find an equation of the tangent line to the graph of f at the given point.
f(x)=27/(x^2+9)
(-3,3/2)
I got y=x/2+3, but I don't think its right.. please help!
f' = -27(2x)/(x^2+9)^2
at x = -3
f'=slope = -27(-6)/(18)^2 = .5
so
y = .5 x + b
put in the point
1.5 = .5 (-3) + b
1.5 = -1.5 + b
b = 3
so
y = .5 x + 3
or
2y = x + 6
To find the equation of the tangent line to the graph of f at the given point, we need to find the slope of the tangent line and use the point-slope form of a linear equation.
First, we need to find the derivative of f(x). Taking the derivative of f(x) will give us the slope of the tangent line at any given point.
To find the derivative, we can use the quotient rule. Let's differentiate f(x)=27/(x^2+9):
f'(x) = [27*(2x)] / (x^2+9)^2
Simplifying this, we get:
f'(x) = (54x) / (x^2+9)^2
Now, let's substitute x = -3 into f'(x) to find the slope of the tangent line at x = -3:
f'(-3) = (54(-3)) / ((-3)^2+9)^2
= (-162) / (9+9)^2
= (-162) / (18)^2
= (-162) / 324
= -1/2
So, the slope of the tangent line to the graph of f at the point (-3, 3/2) is -1/2.
Now, we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y_1 = m(x - x_1)
Let's substitute the values into the equation:
y - 3/2 = (-1/2)(x - (-3))
y - 3/2 = (-1/2)(x + 3)
Simplifying further:
y - 3/2 = (-1/2)x - 3/2
y = (-1/2)x - 3/2 + 3/2
y = (-1/2)x
Therefore, the equation of the tangent line to the graph of f at the point (-3, 3/2) is y = (-1/2)x.
To find the equation of the tangent line to the graph of f at the given point, you can follow these steps:
Step 1: Find the derivative of the function f(x).
Let's find the derivative of f(x) = 27/(x^2+9) using the quotient rule:
f'(x) = [27(x^2+9)' - (x^2+9)(27)'] / (x^2+9)^2
The derivative of the numerator is 27(2x) - 0 (since the derivative of a constant is zero), and the derivative of the denominator is 2(x^2+9).
So, f'(x) = [54x] / (x^2+9)^2.
Step 2: Plug in the x-coordinate of the given point into the derivative to find the slope of the tangent line.
Since the point is (-3, 3/2), plug x = -3 into f'(x):
f'(-3) = [54(-3)] / ((-3)^2+9)^2
= -162 / 36
= -9/2.
Therefore, the slope of the tangent line is -9/2.
Step 3: Use the point-slope form of a line (y - y1 = m(x - x1)) to write the equation of the tangent line.
Using the given point (-3, 3/2) and the slope -9/2, we can write the equation as:
y - (3/2) = (-9/2)(x - (-3))
= (-9/2)(x + 3).
Simplifying further, we have:
y - (3/2) = (-9/2)x - (27/2)
= (-9/2)x - 27/2.
Adding (3/2) to both sides, we get:
y = (-9/2)x - 27/2 + 3/2
= (-9/2)x - 24/2
= (-9/2)x - 12.
So, the equation of the tangent line to the graph of f at the given point (-3, 3/2) is y = (-9/2)x - 12.