How do you find the equation of a line tangent to #y=3x^2-x+4# at x=0?

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Answer 1 · 2016-12-14T22:20:52

#y=4-x #

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=3x^2-x+4# then differentiating wrt #x# gives us:

#dy/dx = 6x-1#

When #x=0 => y=0-0+4# (so #(0,4)# lies on the curve)
and #dy/dx=0-1=-1#

So the tangent we seek passes through #(0,4)# and has gradient #-1# so using #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-4=-1(x-0) #
# :. y-4 = -x#
# :. \ \ \ \ \ \ \ y=4-x #

We can confirm this solution is correct graphically:
enter image source here