Find kutovy tangent coefficient to a function graph in a point with an abscissa x ₀ y=5x³-x+2, x ₀=1 I WILL be VERY GRATEFUL

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The slope of a tangent to a function graph in Huo's point is equal to derivative function in this point. Given: of y (x) =5x³-x+2, x ₀=1. y(x) = 15x ²-1. y(xo) = 15*1²-1 = 14. It also is ug Lowai tangent coefficient to a function graph in Huo's point = 1. the Slope of a tangent to function is a value of derivative function in this point. Y = 5*x³ - x + 2 is GIVEN. Function Y derivative = 15*x² - 1. We calculate value at X = 1. k = Y(1) = 15 - 1 = 14 - slope - the ANSWER In addition. We calculate value of function at X = 1 Y = 5 - 1 + 2 = 6 Coordinate of a point of contact - And (1;6) Equation of a tangent of Y = 14*x + we Find b value - b. b = 6 - 14 =-8 Y tangent equation = 14*x - 8 Graphic decision - as a gift is final.
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