Find the slope of the line tangent to the graph of y = 2x3 – 2x + 57 at (–2, 45).
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The slope = 22.

Step-by-step explanation:

First we find the derivative:

y' = 6x^2 - 2  

This gives the slope of the line tangent to the point (x, y) on the graph of y.

So for the tangent at (-2, 45) the slope is:

6(-2)^2 - 2 =  24 - 2

=  22.

Given that,

The equation of line is y=7/5x+ 6 and that passes through the point (2,-6).

To find,

The equation of line that is perpendicular to the given line.

Solution,

The given line is :

y=7/5x+ 6

The slope of this line = 7/5

For two perpendicular lines, the product of slopes of two lines is :

Equation will be :

y=-5x/7+ b

Now finding the value of b. As it passes through (2,-6). The equation of line will be :

So, the required equation of line is :

y=-5x/7+ (-32/7)