Write the equation of the line tangent to the graph of at .

Question

Write the equation of the line tangent to the graph of  at  . ​

Quizplus · Accepted Answer

To find the equation of the tangent line, we need to find the slope of the curve at the given point. We can take the derivative of the curve and evaluate it at the point to get the slope of the tangent line.

The derivative of the curve 11eb0d3b_987b_9710_9790_c3f2446d4e6d_TB1243_11 is 2x.

When x = 11eb0d3b_987b_9711_9790_e7d71e527c8e_TB1243_11, the slope of the tangent line is 2(11eb0d3b_987b_9711_9790_e7d71e527c8e_TB1243_11) = 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11.

Now we need to find the equation of the line with the slope 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11 that passes through the point (11eb0d3b_987b_9711_9790_e7d71e527c8e_TB1243_11, 11eb0d3b_987b_9710_9790_c3f2446d4e6d_TB1243_11).

Using the point-slope form of the equation of a line, we get:

y - 11eb0d3b_987b_9710_9790_c3f2446d4e6d_TB1243_11 = 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11(x - 11eb0d3b_987b_9711_9790_e7d71e527c8e_TB1243_11)

Simplifying:

y - 11eb0d3b_987b_9710_9790_c3f2446d4e6d_TB1243_11 = 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11x - 11eb0d3b_987b_9716_9790_37ef77a60379_TB1243_11

y = 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11x + 11eb0d3b_987b_9710_9790_c3f2446d4e6d_TB1243_11 - 11eb0d3b_987b_9716_9790_37ef77a60379_TB1243_11

y = 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11x + 11eb0d3b_987b_9710_9790_8d19f2a86384_TB1243_11

Therefore, the equation of the line tangent to the graph of the curve at the point (11eb0d3b_987b_9711_9790_e7d71e527c8e_TB1243_11, 11eb0d3b_987b_9710_9790_c3f2446d4e6d_TB1243_11) is 11eb0d3b_987b_9715_9790_b5b692815ee6_TB1243_11x + 11eb0d3b_987b_9710_9790_8d19f2a86384_TB1243_11.

Write the Equation of the Line Tangent to the Graph