Find the slope m and an equation of the tangent line to the graph of the function f at the specified point.
f(x)=x^2/x+4 ; (3,9/7)
f ' (x) = ((x+4)(2x) - 2x(1) )/(x+4)^2
= (2x^2 + 6x)/(x+4)^2
when x = 3
slope = 36/49
so y = (36/49)x + b
at (3,9/7)
9/7 = (36/49)(3) + b
b = 9/7 - 108/49
= -45/49
y = (36/49)x - 45/49 or (36x - 45)/49
check my arithmetic
To find the slope (m) and equation of the tangent line to the graph of the function (f) at a specified point (3, 9/7), we need to use the concept of derivative.
1. Begin by finding the derivative of the function (f'(x)). To do this, we can apply the quotient rule. Recall that the quotient rule states: (d/dx)(u/v) = (v * du/dx - u * dv/dx) / (v^2)
In our case, u = x^2 and v = x + 4. Therefore, we have:
f'(x) = [(x + 4) * (d/dx)(x^2) - x^2 * (d/dx)(x + 4)] / (x + 4)^2
Applying the power rule, we differentiate x^2 as 2x, and the derivative of x + 4 is simply 1. Simplifying the expression:
f'(x) = [(x + 4) * 2x - x^2 * 1] / (x + 4)^2
= (2x(x + 4) - x^2) / (x + 4)^2
2. Now we need to find the slope (m) of the tangent line at the point (3, 9/7). We substitute x = 3 into the derivative expression we found in Step 1:
m = f'(3) = [2(3)(3 + 4) - 3^2] / (3 + 4)^2
= [2(3)(7) - 9] / (7)^2
= (42 - 9) / 49
= 33 / 49
Therefore, the slope (m) of the tangent line is 33/49.
3. To find the equation of the tangent line, we use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the given point on the graph.
Using the point (3, 9/7) and the slope (m) we found in Step 2, we can plug these values into the equation:
y - (9/7) = (33/49)(x - 3)
Expanding and rearranging the equation, we get:
y = (33/49)x - (99/49) + (9/7)
= (33/49)x - (99/49) + (63/49)
= (33/49)x - (36/49)
Therefore, the equation of the tangent line to the graph of the function at the point (3, 9/7) is y = (33/49)x - (36/49).