For the given function, find the equation of the secant line through the points

Question

For the given function, find the equation of the secant line through the points where x has the given values and the equation of the tangent line when x has the first value. y = f(x) = x2 + x; x = - 4, x = -1 The equation of the secant line is y = (Simplify your answer. Type your answer in slope-intercept form.)

Explanation / Answer

a. The two points are (-4,12) and (-1,0) The secant line connecting those two points is y=mx+b with m=(12-0)/(-4+1)=-4 y=-4x+b finding b: use the point (-1,0) 0=4+b ==> b = -4 The secant line is y=-4x-4 b. The tangent line has slope which is the derivative of the function at the point (-4,12) f'(x)=2x+1; f'(-4)=-8+1=-7 So the tangent line is y=-7x+b finding b: use the point (-4,12) (we have only one point now), 12=-7*(-4)+b ==> b=12-28=-16 The tangent line is y=-7x-16

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