PLEASE HELP Find the slope and equation of the tangent to each of the following curves at the given point:

Step-by-step explanation:

To find the slope of the tangent at a point we will find the derivative of equation at that point.

a). y = x² - 10x

y' = 2x - 10

At (x = 2),

y' = 2(2) - 10

y' = 4 - 10

y' = -6

From the given equation,

y = 2² - 10

y = -6

Therefore, y-coordinate of the point is y = -6

Equation of the tangent at (2, -6) having slope = -6

y - 2 = -6(x - 2)  

y - 2 = -6x + 12

y = -6x + 14

b). y =

At x =

y =

y =

y =

Now we have to find the equation of a tangent at

y' = 2x -

At x =

y' =

y' = 1 - 8

y' = -7

Therefore, equation of the tangent at will be,

y = -7x +

y = -7x +

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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