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I need help finding the equation for a tangent line

I need help finding the equation for a tangent line-example-1

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To answer this question we will use the following slope-point formula for the equation of a line:


y-y_1=m(x-x_1)\text{.}

Recall that f'(x) is the slope of the tangent line to the graph at (x,f(x)).

Therefore, the tangent line of y=f(x) at x=2 has a slope of -4 and passes through (-2,-7).

Then, using the slope-point formula for the equation of a line we get:


y-(-7)=-4(x-(-2))\text{.}

Simplifying the above result we get:


\begin{gathered} y+7=-4(x+2), \\ y+7=-4x-8. \end{gathered}

Subtracting 7 from the above equation we get:


\begin{gathered} y+7-7=-4x-8-7, \\ y=-4x-15. \end{gathered}

Answer:


y=-4x-15.

User Sacha Barber
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