Question

Find the derivative of the function y = 2x^2 - 13x + 5 and use it to find the equation of the line tangeant to the curve at x = 3

Answer 1

The derivative of the function y = 2x^2 - 13x + 5 is y' = 4x - 13. The equation of the line tangent to the curve at x = 3 is y = -x - 13.

Step-by-step explanation:

To find the derivative of the function y = 2x^2 - 13x + 5, we use the power rule of differentiation. This gives us the derivative y' = 4x - 13. To find the equation of the line tangent to the curve at x = 3, we first evaluate the derivative at this point to get the slope, which is y'(3) = 4(3) - 13 = 12 - 13 = -1. Then, we plug in x = 3 into the original function to get the y-coordinate of the tangent point, which is y(3) = 2(3)^2 - 13(3) + 5 = 18 - 39 + 5 = -16. Therefore, the point of tangency is (3, -16). Using the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point of tangency, we can write the equation of the tangent line: y - (-16) = -1(x - 3), which simplifies to y = -x - 13.

Answer 2

y = 2x^2 - 13x + 5
dy/dx = 4x - 13

At x = 3, dy/dx = 4(3) - 13 = 12 - 13 = -1
and y = 2(3)^2 - 13(3) + 5 = 2(9) - 39 + 5 = 18 - 34 = -16

The tangent line at x = 3, passes through the point (3, -16) and has a slope of -1
y - (-16) = -1(x - 3)
y + 16 = -x + 3
y = -x - 13 is the required equation.