a. For the following function, find f (a).
b. Determine an equation of the line tangent to the graph of f at (a,f(a)) for the given value of a.
f(x)= 9/(x+4) ,a=2
f'(x) = -9/(x+4)^2
at x=2, f'(2) = -1/4
f(2) = 3/2
so now you have a point (2,3/2) and a slope (-1/4) so the point-slope equation for the tangent line is
y - 3/2 = -1/4 (x-2)
a. To find f(a) for the given function f(x) = 9/(x+4) and a = 2, let's substitute x with a in the function:
f(a) = 9/(a+4)
Therefore, f(2) = 9/(2+4) = 9/6 = 3/2.
b. Now, let's determine the equation of the tangent line to the graph of f at (a, f(a)) for the given value of a = 2. To do that, we need to find the slope of the tangent line.
We find the derivative of f(x) = 9/(x+4) using exponent rules and the chain rule:
f'(x) = -9/(x+4)²
Now, let's substitute x with a in the derivative:
f'(a) = -9/(a+4)²
Using the point-slope form of a line, the equation of the tangent line is y - f(a) = f'(a)(x - a). Substituting the values, we have:
y - (3/2) = -9/(a+4)² (x - 2)
So, the equation of the tangent line to the graph of f when a = 2 is y - 3/2 = -9/36 (x - 2). Simplifying it further might be a tangent-ial task!
a. To find f(a), substitute 2 for x in the function f(x):
f(a) = f(2) = 9/(2+4) = 9/6 = 3/2
Therefore, f(a) = 3/2.
b. To determine the equation of the tangent line to the graph of f at (a, f(a)) for the given value of a, we need to find the derivative of f(x) and use it to find the slope of the tangent line.
First, let's find the derivative of f(x) using the quotient rule:
f'(x) = [(d/dx)(9)]*(x+4) - 9*(d/dx)(x+4)] / (x+4)^2
= 0*x - 9*1 / (x+4)^2
= -9/(x+4)^2
Next, substitute a=2 into the derivative to find the slope of the tangent line at x=a:
f'(a) = -9/(a+4)^2
= -9/(2+4)^2
= -9/36
= -1/4
So, the slope of the tangent line at x=a is -1/4.
Now, we can use the point-slope form of a line to find the equation of the tangent line. We have the point (a, f(a)) = (2, 3/2) and the slope m = -1/4.
Using the point-slope form, the equation of the tangent line is:
y - y1 = m(x - x1)
y - (3/2) = (-1/4)(x - 2)
y - 3/2 = -1/4x + 1/2
y = -1/4x + 1/2 + 3/2
y = -1/4x + 2
Therefore, the equation of the line tangent to the graph of f at (a,f(a)) for the given value of a=2 is y = -1/4x + 2.
To find f(a) for the function f(x) = 9/(x+4) when a = 2, we simply substitute the value of a into the function.
a. Finding f(a):
f(2) = 9/(2+4) = 9/6 = 3/2
Therefore, f(a) = 3/2 when a = 2.
b. To determine the equation of the line tangent to the graph of f at (a, f(a)), we need to find the derivative of f(x) first. Let's do that.
Finding the derivative of f(x):
To find the derivative, we need to use the quotient rule. The quotient rule states that for a function h(x) = g(x)/k(x), the derivative of h(x) is given by h'(x) = [g'(x)k(x) - k'(x)g(x)] / k^2(x).
In our case, g(x) = 9 and k(x) = (x+4).
Using the quotient rule, we have:
f'(x) = [9 * 1 - 0 * (x+4)] / (x+4)^2
= 9 / (x+4)^2
Now, let's substitute a = 2 into the derivative to find the slope of the tangent line at (a, f(a)).
Slope of the tangent line:
f'(a) = 9 / (a+4)^2
= 9 / (2+4)^2
= 9 / (6)^2
= 9 / 36
= 1/4
Therefore, the slope of the tangent line at (2, 3/2) is 1/4.
Now, we have the slope of the tangent line and a point on the line (2, 3/2). We can use the point-slope form of the equation of a line to find the equation.
Equation of the tangent line:
Using the point-slope form y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point, we substitute m = 1/4 and (x1, y1) = (2, 3/2).
y - 3/2 = 1/4(x - 2)
To simplify, we can multiply both sides by 4 to get rid of the fraction:
4y - 6 = x - 2
Rearranging the equation:
x - 4y = -4
Therefore, the equation of the tangent line to the graph of f at (2, 3/2) is x - 4y = -4.