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Find an equation of the tangent line to the curve at the given point.y=7 x² -x³,(1,6) [ y=_] Illustrate by graphing the curve and the tangent line on the same screen.

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Final answer:

Calculate the derivative of y = 7x² - x³ to find the slope of the tangent, substitute x = 1 to get the slope at the given point (1,6). Use the slope and the point to write the equation of the tangent line using the point-slope form. The equation is y = 11x - 5.

Step-by-step explanation:

To find an equation of the tangent line to the curve at the given point (1,6) for the function y = 7x² - x³, you first need to calculate the derivative of the function, which represents the slope of the tangent line at any point x. The function's derivative is dy/dx = 14x - 3x². Substituting x = 1 gives the slope m = 14(1) - 3(1)² = 11. Now, we have a point and a slope, so we can use the point-slope form to write the equation of the tangent line: y - y₁ = m(x - x₁). Substituting the slope we found and the point (1,6), we get y - 6 = 11(x -1). Simplifying, the equation of the tangent line is y = 11x - 5. To illustrate this by graphing, imagine a curve representing y = 7x² - x³ and a straight line with the equation y = 11x - 5. The curve will be parabolic opening downwards (since the coefficient of is negative), and the tangent line will intersect this curve precisely at the point (1,6), touching without crossing at that exact point.

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