[tex]\boxed{y+9=7(x+1)}[/tex]
Explanation:
The Point-Slope form of the equation of a line is given by:
[tex]y-y_{0}=m(x-x_{0}) \\ \\ \\ Where: \\ \\ m:Slope \\ \\ (x_{0},y_{0}):A \ point[/tex]
To find the equation of the line tangent to [tex]y=-3x^2+x-5 \ at \ x=-1[/tex], we need to take the derivative:
[tex]\frac{dy}{dx}=-6x+ 1[/tex]
The slope of the line tangent to [tex]y=-3x^2+x-5[/tex] at [tex]x=-1[/tex] can be found evaluationg [tex]x=1[/tex] in the derivative:
[tex]\frac{dy}{dx}\Bigr\rvert_{x=-1}=-6x+ 1 \\ \\ \\ \frac{dy}{dx}\Bigr\rvert_{x=-1}=-6(-1)+ 1 \\ \\ \\ m=\frac{dy}{dx}\Bigr\rvert_{x=-1}=7[/tex]
Finding the point:
[tex]x_{0}=-1 \\ \\ Then: \\ \\ y_{0}=-3(-1)^2-1-5=-9 \\ \\ (x_{0},y_{0})=(-1,-9)[/tex]
Finally, our line is:
[tex]y-(-9)=7(x-(-1)) \\ \\ \boxed{y+9=7(x+1)}[/tex]
The representation of this problem is shown below.
Learn more:
Derivative of functions: https://brainly.com/question/2142425
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