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

Question

Scarlett Allison · Accepted Answer

# y=4x-9#

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

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

#dy/dx = 2x-2#

When #x=3 => y=9-6=3# (so #(3,3)# lies on the curve)
and #dy/dx=6-2=4#

So the tangent we seek passes through #(3,3)# and has gradient #4# so using #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-3=4(x-3) #
# :. y-3=4x-12#
# :. \ \ \ \ \ \ \ y=4x-9#

Adam Adkins · Answer

undefined To find the equation of a line tangent to a curve at a given point, we need to determine the slope of the curve at that point. To do this, we take the derivative of the curve's equation with respect to x.

The derivative of y = x^2 - 2x is dy/dx = 2x - 2.

Next, we substitute the x-coordinate of the given point (3) into the derivative to find the slope at that point.

dy/dx = 2(3) - 2 = 4.

Therefore, the slope of the curve at (3,3) is 4.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

y - 3 = 4(x - 3).

Simplifying, we get y = 4x - 9.

Thus, the equation of the line tangent to y = x^2 - 2x at (3,3) is y = 4x - 9.