g(6) = -2

g'(6) = 5

The equation of the tangent line is given by:

y-y1 = m (x-x1) such that: (x1,y1) is any point of the graph of g(x) , and m is the slope.

Given that g(6)= -2

==> Then, the point (6,-2) is on the graph.

==>...

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g(6) = -2

g'(6) = 5

The equation of the tangent line is given by:

y-y1 = m (x-x1) such that: (x1,y1) is any point of the graph of g(x) , and m is the slope.

Given that g(6)= -2

==> Then, the point (6,-2) is on the graph.

==> Also, we know that the slope is the derivative at the point of tendency which is x= 6

Then, the slope is g'(6)= 5

==> m= 5

Now we will plug in the values.

==> y-(-2) = 5(x-6)

==> y+2 = 5x - 30

==> y= 5x - 32

**Then, the equation of the tangent line is y=5x-32**