92.7k views
3 votes
For f(x)=1/5+x^2, the slope of the graph of y=f(x) is known to be -1/18 at the point with x-coordinate 1. Find the equation of the tangent line at the point.

User Simcha
by
8.7k points

1 Answer

1 vote

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.

User Captaintom
by
8.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories