Question

Find the equation of the tangent line and the normal line through the point (-2, 8) for the function y = 3x^2 + 7x + 10.

A) Tangent Line: y = 3x + 14, Normal Line: y = -3x + 22

B) Tangent Line: y = 14x - 6, Normal Line: y = -14x + 22

C) Tangent Line: y = 6x + 14, Normal Line: y = -6x + 22

D) Tangent Line: y = 22x - 14, Normal Line: y = -22x + 14

Answer 1

To find the equation of the tangent line and the normal line through the point (-2, 8) for the function y = 3x^2 + 7x + 10, we first find the derivative of the function. Then, we substitute the x-coordinate of the point into the derivative to find the slope of the tangent line. The equation of the tangent line is y = -5x + 18, and the equation of the normal line is y = 1/5x + 8/5.

Step-by-step explanation:

To find the equation of the tangent line and the normal line through the point (-2, 8) for the function y = 3x^2 + 7x + 10, we need to find the derivative of the function. The derivative of y = 3x^2 + 7x + 10 is y' = 6x + 7.

Then, we can find the slope of the tangent line at (-2, 8) by substituting x = -2 into the derivative:

m = 6(-2) + 7 = -5.

Using the point-slope form of a line, we can write the equation of the tangent line as y - 8 = -5(x + 2), which simplifies to y = -5x + 18.

The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is 1/5.

Using the point-slope form again, we can write the equation of the normal line as y - 8 = 1/5(x + 2), which simplifies to y = 1/5x + 8/5.