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# What is the tangent equation for the function f(x)=(2x+3)^5, where x=-1?

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To get the equation of the tangent line you need to evaluate the derivative at x=-1 to find the slope m, and evaluate the function at x=-1 to find the intercept for y=mx+b.  The function (2x+3)^5 at x=-1 evaluates to 1.

Let u be 2x+3. The derivative of u^n is nu^(n-1) du/dx which is 5(2x+3)^4 times 2, which at x=-1 comes out to 5(1)(2) = 10.  If you didn't know that derivativative method then you could expand (2x+3)^5 and find the derivative of each term.

To find y=mx+b, we know m=10, and 1=10(-1)+b so b=11 and the tangent is y=10x+11.

It's been too long for me.

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