How do you find the equation of a line tangent to the function #y=x-2x^2+3# at x=2?

Question

Emilia Aguilar · Accepted Answer

# y = -7x+11 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

so If # y = x-2x^2+3 # then differentiating wrt #x# gives us:

# dy/dx = 1-4x #

When #x=2 => y=2-8+3=-3# (so #(2,-3)# lies on the curve)
and # dy/dx = 1-8=-7#

So the tangent passes through #(2,-3)# and has gradient #-7#, so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-(-3) = -7(x-2) #
# :. y+3 = -7x+14#
# :. \ \ \ \ \ \ \ y = -7x+11 #

We can confirm this solution is correct graphically:

Adam Adkins · Answer

undefined To find the equation of a line tangent to a function at a specific point, you need to follow these steps:

1. Find the derivative of the function.
2. Evaluate the derivative at the given x-value to find the slope of the tangent line.
3. Use the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope, to write the equation of the tangent line.

Let's apply these steps to the given function y = x - 2x^2 + 3 at x = 2:

1. Find the derivative of the function:
   y' = 1 - 4x

2. Evaluate the derivative at x = 2:
   y'(2) = 1 - 4(2) = -7

3. Use the point-slope form with the point (2, y(2)):
   y - y₁ = m(x - x₁)
   y - y(2) = -7(x - 2)

Simplifying the equation:
   y - (2 - 2(2)^2 + 3) = -7(x - 2)
   y - (-5) = -7(x - 2)
   y + 5 = -7x + 14

The equation of the line tangent to the function y = x - 2x^2 + 3 at x = 2 is:
   y = -7x + 9