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

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Answer 1 · 2018-04-15T11:00:23

#y=x-7#

Let #y=f(x)=x^2-5x+2#

At #x=3,y=3^2-5*3+2#

#=9-15+2#

#=-6+2#

#=-4#

So, the coordinate is at #(3,-4)#.

We first need to find the slope of the tangent line at the point by differentiating #f(x)#, and plugging in #x=3# there.

#:.f'(x)=2x-5#

At #x=3#, #f'(x)=f'(3)=2*3-5#

#=6-5#

#=1#

So, the slope of the tangent line there will be #1#.

Now, we use the point-slope formula to figure out the equation of the line, that is:

#y-y_0=m(x-x_0)#

where #m# is the slope of the line, #(x_0,y_0)# are the original coordinates.

And so,

#y-(-4)=1(x-3)#

#y+4=x-3#

#y=x-3-4#

#y=x-7#

A graph shows us that it's true:

desmos.com

Answer 2 · 2018-04-15T11:07:16

#y = x - 7#

#y=x^2-5x+2#
#y' = 2x - 5#

At #x=3 :#

#y' = 2x - 5#
#y' = 6 - 5#
#y' = 1#

#y = 3^2 - 5 xx 3 + 2#
#y = -4#

#y' = 1, (3, -4)#
#y - (-4) = 1(x - 3)#
#y = x - 7#