Find the slope of the tangent to the curve $y = 3x^2 - 2x + 1$ at $x = 1$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the slope of the tangent line to the curve $ y = 3x^2 - 2x + 1 $ at $ x = 1 $, we need to compute the derivative of $ y $ with respect to $ x $. ‖ ‖ step 2 ⋮ The derivative of $ y $ with respect to $ x $ is $ \frac{dy}{dx} = 6x - 2 $. ‖ ‖ step 3 ⋮ Evaluate the derivative at $ x = 1 $ to find the slope of the tangent line at that point: $ \frac{dy}{dx} \bigg|_{x=1} = 6(1) - 2 $. ‖ ‖ step 4 ⋮ Simplifying gives us the slope: $ 6 - 2 = 4 $. ‖ ∻Answer∻ ⚹ The slope of the tangent to the curve at $ x = 1 $ is 4. ⚹ ∻Key Concept∻ ⚹Finding the slope of a tangent line to a curve at a given point⚹ ∻Explanation∻ ⚹The slope of the tangent line to a curve at a particular point is found by evaluating the derivative of the curve's equation at that point.⚹◊What is the formula to find the slope of a tangent line to a curve at a specific point?⍭ Generate me a similar question◊

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