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For the function f(x)=8x³-2x²+2x, step 2 of 2: Find the equation of the tangent line at x=2?

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Final answer:

To find the equation of the tangent line at x=2 for the function f(x) = 8x³ - 2x² + 2x, we need to find the slope of the tangent line at that point and use the point-slope form of a linear equation.

Step-by-step explanation:

To find the equation of the tangent line at x=2 for the function f(x) = 8x³ - 2x² + 2x, we need to find the slope of the tangent line at that point and use the point-slope form of a linear equation.

To find the slope, we need to find the derivative of the function and evaluate it at x=2. Taking the derivative, we get f'(x) = 24x² - 4x + 2. Evaluating at x=2, we get f'(2) = 24(2)² - 4(2) + 2 = 88.

So the slope of the tangent line at x=2 is 88. Now we can use the point-slope form, which is y - y₁ = m(x - x₁), where (x₁, y₁) is the point (2, f(2)). Substituting the values, we get y - f(2) = 88(x - 2). Simplifying, we get the equation of the tangent line as y = 88x - 172.

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