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Find the coordinates of the points on the graph of #y=3x^2-2x# at which tangent line is parallel to the line #y=10x#?

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Jim G. Share
Feb 18, 2018

Answer:

#(2,8)#

Explanation:

#•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"#

#y=3x^2-2x#

#rArrdy/dx=6x-2#

#y=10xtom=10#

#rArr6x-2=10rArrx=2#

#x=2toy=12-4=8#

#rArr"coordinates of point "=(2,8)#

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Feb 18, 2018

Answer:

#(2,8)#

Explanation:

Two lines are parallel if they share the same slope.

Any line parallel to #y=10x# will have the slope 10.

Now, we have: #y=3x^2-2x#

We try to find #dy/dx#

=>#d/dx(y)=d/dx(3x^2-2x)#

We use the power rule:

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.

#dy/dx=3*2x^(2-1)-2*1x^(1-1)#

=>#dy/dx=6x-2#

Now, we set #dy/dx=10#

=>#10=6x-2#

=>#12=6x#

=>#2=x#

We plug this value in to get our #y#, or #f(x)#.

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

=>#12-4=y#

=>#8=y#

Therefore, the point is at #(2,8)#

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