How do you find the equation of tangent line to the curve #y=2x^2+4x-3# at the point (1,3)?

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Answer 1 · 2016-12-09T13:56:23

# y=8x-5#

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

#dy/dx = 4x+4#

When #x=1 => y=2+4-3=3# (so #(1,3)# lies on the curve)
and #dy/dx=4+4=8#

So the tangent we seek passes through #(1,3)# ad has gradient 8 so using #y-y_1=m(x-x_1)# the equation we seek is;

# y-3=8(x-1) #
# :. y-3=8x-8#
# :. y=8x-5#

We can confirm this graphically:

enter image source here