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Given y = -1 x^2 + 4 x + (-1), find the equation of the tangent line that is horizontal to the curve. The slope-intercept form of the equation of the tangent line that is horizontal to the curve is y =.
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dy/dx=0
dy/dx=-2x+4=0 x=2 y=3 y-y1=m(x-x1)
horizontal to the curve means m = 0 in a parabola this is at the stationary point so dy/dx = -2x + 4 dy/dx = 0 when x = 2 point (2, 3) then y = 3 is the horizontal tangent
thanks this was correct
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