The curve C has the equation y = 2x^2 -11x + 13. Find the equation of the tangent to C at the point P (2, -1).

The first step is to differentiate the equation of the curve in order to find the gradient of the tangent at the curve. Remember that when differentiating polynomials, we multiply the index of the variable x, by its coefficient, then subtract 1 from the index. In addition, remember that x0 = 1.

In this case, dy/dx = 4x - 11.

Now if we plug in the x-coordinate of P (2) into dy/dx, we will get the gradient of the tangent to the curve at P.

dy/dx = 4(2) - 11

dy/dx = 8 - 11

dy/dx = -3.

Now we find the equation of the tangent using the formula for the equation of a straight line, and plugging in the coordinates of P:

y - y= m(x - x1)

y - (-1) = -3(x - 2)

y + 1 = -3x +6

3x + y - 5 = 0.

This is the equation of the tangent to the curve C at P.

Answered by James Y. Maths tutor

15328 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is dy/dx when y=ln(6x)?


A circle C with centre at the point (2, –1) passes through the point A at (4, –5). Find an equation for the circle C.


How do I find the turning points of a curve?


Consider the functions f(x) = −x^3 + 2x^2 + 3x and g(x) = −x^3 + 3x^2 − x + 3. (a) Find df/dx (x) and hence show that f(x) has turning points at when x = 2 /3 ± √ 13/ 3 . [5] (b) Find the points where f(x) and g(x) intersect. [4]


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences