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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].
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