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 at a specific point, we need to calculate the derivative of the function. ‖ ‖ step 2 ⋮ The derivative of the function $ y = 3x^2 - 2x + 1 $ is found using the power rule. ‖ ‖ step 3 ⋮ Applying the power rule, the derivative is $ \frac{dy}{dx} = 6x - 2 $. ‖ ‖ step 4 ⋮ To find the slope at $ x = 1 $, we substitute $ x = 1 $ into the derivative. ‖ ‖ step 5 ⋮ Substituting $ x = 1 $ gives us $ \frac{dy}{dx} = 6(1) - 2 = 4 $. ‖ ∻Answer∻ ⚹ The slope of the tangent to the curve at $ x = 1 $ is 4. ⚹ ∻Key Concept∻ ⚹Derivative as Slope of Tangent⚹ ∻Explanation∻ ⚹The derivative of a function at a point gives the slope of the tangent line to the curve at that point.⚹◊What is the equation of the tangent line to the curve $y = 3x^2 - 2x + 1$ at $x = 1$?⍭ Generate me a similar question◊

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