# How do you find the slope of the secant lines of f(x)=x^3-12x+1 through the points: -3 and 3?

slope $= - 3$

#### Explanation:

Given : $f \left(x\right) = {x}^{3} - 12 x + 1$

when $x = - 3$ find $f \left(- 3\right)$

$f \left(- 3\right) = {\left(- 3\right)}^{3} - 12 \cdot \left(- 3\right) + 1 = - 27 + 36 + 1 = 10$

We have a point $\left(- 3 , 10\right)$
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when $x = 3$ find $f \left(3\right)$

$f \left(3\right) = {\left(3\right)}^{3} - 12 \cdot \left(3\right) + 1 = 27 - 36 + 1 = - 8$

We have a point $\left(3 , - 8\right)$
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Solve the slope
Slope$= \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{10 - - 8}{- 3 - 3} = \frac{18}{-} 6 = - 3$

God bless....I hope the explanation is useful.