**Q: Find the point(s) on the graph of the function f(x) = (x² + 6)(x – 5) where the slope of the tangent line is -2.**

A: The derivative measures the slope of the tangent line. So, I must first take the derivative of f(x):

f(x) = (x² + 6)(x – 5)

To take the derivative, use the product rule (which tells us the derivative of f*g is f ‘ * g + g ‘ * f)

f = (x² + 6) g = (x – 5)

f ‘ = 2x g ‘ = 1

So, f ‘ (x) = 2x(x – 5) + 1(x² + 6)

= 2x² – 10x + x² + 6 = 3x² – 10x + 6

f ‘ (x) = 3x² – 10x + 6

Now, we want the slope to equal -2… Remember… The derivative is the slope

-2 = 3x² – 10x + 6

So, we just need to solve for x:

0 = 3x² – 10x + 8 [I added 2 to both sides]

0 = (3x – 4)(x – 2)

0 = (3x – 4) **or **0 = (x – 2)

x = 4/3 **or **x = 2

Now that we have the x values, we must find the y values that go with each x (since we are asked to find **points**). Plug in each x (separately) to the original equation to find y:

f(4/3) = ((4/3)² + 6)(4/3 – 5) = (16/9 + 6)(4/3 – 5) = 70/9 * -11/3 = -770/27

The first point is **(4/3, -770/27)**

f(2) = (2² + 6)(2 – 5) = (4 + 6)(-3) = 10*-3 = -30

The second point is **(2, -30)**

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