find the slope m and an equation of the tangent line to the graph of the function f at the specified point.
original: x^2/x+10 point: (3,9/13)
After I found the derivative: (x+10)(2x)-(x^2)(1)/((x+10))^2
How do I get the slope? Should I put 3 into the derivative equation?
To find the slope (m) of the tangent line to the graph of the function f at a specific point, you need to evaluate the derivative of the function at that point.
In this case, you have already found the derivative of the function f to be:
f'(x) = (x+10)(2x) - (x^2)(1) / (x+10)^2
To find the slope at the point (3, 9/13), you can substitute the x-coordinate (3) into the derivative equation:
f'(3) = (3+10)(2(3)) - (3^2)(1) / (3+10)^2
Simplifying the expression, you get:
f'(3) = (13)(6) - (9) / (13)^2
Calculate the numerator and denominator separately:
Numerator: (13)(6) - (9) = 78 - 9 = 69
Denominator: (13)^2 = 169
So, the slope of the tangent line at the point (3, 9/13) is:
m = f'(3) = 69/169
Now that you have the slope (m), you can proceed to find the equation of the tangent line using the point-slope form:
y - y1 = m(x - x1)
Using the point (3, 9/13), and the slope 69/169, you can substitute these values into the equation to get the final equation of the tangent line to the graph of the function f.