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Given f(x)=x^3-2, find the equation of the secant line passing through (-4,f(-4)) and (2,f(2))

Sagot :

Given:

The function is

[tex]f(x)=x^3-2[/tex]

The secant line passing through (-4,f(-4)) and (2,f(2)).

To find:

The equation of the secant line.

Solution:

We have,

[tex]f(x)=x^3-2[/tex]

At x=-4,

[tex]f(-4)=(-4)^3-2[/tex]

[tex]f(-4)=-64-2[/tex]

[tex]f(-4)=-66[/tex]

At x=2,

[tex]f(2)=(2)^3-2[/tex]

[tex]f(2)=8-2[/tex]

[tex]f(2)=6[/tex]

The secant line passes through the points (-4,-66) and (2,6). So, the equation of the secant line is

[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

[tex]y-(-66)=\dfrac{6-(-66)}{2-(-4)}(x-(-4))[/tex]

[tex]y+66=\dfrac{6+66}{2+4}(x+4)[/tex]

[tex]y+66=\dfrac{72}{6}(x+4)[/tex]

On further simplification, we get

[tex]y+66=12(x+4)[/tex]

[tex]y+66=12x+48[/tex]

[tex]y=12x+48-66[/tex]

[tex]y=12x-18[/tex]

Therefor, the equation of the secant line is [tex]y=12x-18[/tex].