How do you find the equation of a line tangent to the function #y=x^2+3x-1# at x=0?

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Answer 1 · 2016-11-17T13:13:37

#y=3x-3#

The equation of a straight line with given gradient #m# and one known point #(x_1,y_1)# is:

#y-y_1=m(x-x_1)#

1) for #y=x^2+3x-1 # at #x=0#

#y(0)=0+0-1=-1#

#(x_1,y_1)=(0,-1)#

2) the gradient of the tangent at the given point is found by differentiating.

#y=x^2+3x-1 #

#(dy)/(dx)=2x+3#

#((dy)/(dx))_(x=0)=2xx0+3=3#

3) use #y-y_1=m(x-x_1)#

#y-(-1)=3(x-0)#

#y+1=3x#

#:.y=3x-1#