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 $y' = \frac{dy}{dx} = 6x - 2$. ‖ ‖ step 4 ⋮ To find the slope at $x = 1$, we substitute $1$ into the derivative to get $y'(1) = 6(1) - 2$. ‖ ‖ step 5 ⋮ Simplifying gives us the slope at $x = 1$, which is $y'(1) = 6 - 2 = 4$. ‖ ∻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 formula for finding the slope of a tangent line to a curve at a specific point?⍭ Generate me a similar question◊

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