Explanation:
To find the equation of the tangent line at the point with x-coordinate 1, we need to determine both the slope and the y-coordinate of the point on the graph of the function f(x) = 1/5 + x^2.
Given that the slope of the graph at the point (1, f(1)) is -1/18, we can differentiate the function f(x) with respect to x to find the derivative.
f'(x) = d/dx (1/5 + x^2)
= 2x
Next, we can evaluate the derivative at x = 1 to find the slope at that point:
f'(1) = 2(1)
= 2
Since the slope of the graph at the point (1, f(1)) is -1/18, we can set up the following equation:
2 = -1/18
Solving for the y-coordinate:
f(1) = 1/5 + 1^2
= 1/5 + 1
= 6/5
Therefore, the y-coordinate of the point on the graph is 6/5.
Now that we have both the slope (-1/18) and the y-coordinate (6/5) of the point (1, f(1)), we can use the point-slope form of the equation of a line to find the equation of the tangent line.
y - y1 = m(x - x1)
Substituting the values:
y - (6/5) = (-1/18)(x - 1)
Simplifying:
y - 6/5 = (-1/18)x + 1/18
Bringing y to the left side:
y = (-1/18)x + 1/18 + 6/5
y = (-1/18)x + (1/18 + 36/18)
y = (-1/18)x + 37/18
Therefore, the equation of the tangent line to the graph of f(x) = 1/5 + x^2 at the point (1, 6/5) is y = (-1/18)x + 37/18.