How do you find an equation of the tangent line to the parabola #y=x^2-2x+7# at the point (3,10)?

1 Answer
Steve M
Jan 5, 2017

# y=4x-2#

Explanation:

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent, so the product of their gradients is #-1#

so If #y=x^2-2x+7# then differentiating wrt #x# gives us:

#dy/dx = 2x-2#

When #x=3 => y=9-6+7=10# (so #(3,10)# lies on the curve)
and #dy/dx=6-2=4#

So the tangent we seek passes through #(3,10)# ad has gradient #4# so using #y-y_1=m(x-x_1)# the equation we seek is;

# y-10=4(x-3) #
# :. y-10=4x-12#
# :. y=4x-2#

We can confirm this graphically:
enter image source here

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