Mark M. answered 07/04/15
Tutor
4.9
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
y = x2-3x-4 = (x-4)(x+1) = 0 x = -1 or x = 4
Let A = (-1,0) and B = (4,0)
y' = 2x-3 Slope of tangent at A = 2(-1)-3 = -5
Slope of tangent at B = 2(4)-3 = 5
Equation of tangent line at A: y - 0 = -5(x + 1) y = -5x -5
Equation of tangent line at B: y - 0 = 5(x - 4) y = 5x - 20
The tangent lines intersect when -5x-5 = 5x-20
-10x = -15
x = 1.5
and y = -5(1.5)-5
= -12.5
So, T = (1.5, -12.5)
Equation of normal line at A: y - 0=(1/5)(x + 1) y = (1/5)x + 1/5
Equation of normal line at B: y - 0 =(-1/5)(x - 4) y = (-1/5)x + 4/5
The normal lines intersect when (1/5)x+1/5 = (-1/5)x + 4/5
(2/5)x = 3/5
x = 1.5
and y = (-1/5)(1.5) + 4/5
= 0.5
So, N = (1.5, 0.5)
Area of quadrilateral ATBN = Area of ΔABT + Area of ΔABN
= ½(5)(12.5) + ½(5)(0.5)
= 32.5