How do you find the equation of the tangent line to the curve #3x^2-5x+2# at x=3?

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Answer 1 · 2016-06-12T01:48:48

First, let's find the value of #y# for #(3, y)# to be on the graph of the function.

#y = 3(3)^2 - 5(3) + 2#

#y = 27 - 15 + 2#

#y = 14#

Next, differentiate using the power rule. Let your function be #f(x)#, then:

#f'(x) = 6x - 5#

Now, plugging in our value of x to find the slope:

#f'(3) = 6(3) - 5#

#f'(3) = 13#

#:.# The slope of the tangent is #13#. Now we know a point on the original function #(3, 14)# and the slope of the tangent, 13.

We will use point slope form to determine the equation of the tangent.

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

#y - 14 = 13(x - 3)#

#y - 14 = 13x - 39#

#y = 13x - 25#

#:.# The equation of the tangent is #y = 13x - 25#.

Hopefully this helps!