Question
In Exercises $5-10$ , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together.$$y=(x-1)^{2}+1, \quad(1,1)$$
Step 1
The derivative of a function gives us the slope of the tangent line at any point on the curve. The given function is $y=(x-1)^{2}+1$. Using the chain rule, the derivative of this function is: $$ y' = 2(x-1) \cdot 1 = 2(x-1) $$ Show more…
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In Exercises $5-10$ , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$ y=(x-1)^{2}+1, \quad(1,1) $$
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In Exercises $5-10$ , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$ y=\frac{1}{x^{3}}, \quad\left(-2,-\frac{1}{8}\right) $$
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