Answer:
P is (2,4) and Q is (50,0).
Step-by-step explanation:
Where the 2 curves intersect at point P :
4e^(x^2 - x - 2) = 8e^(x^2 - 2x - ln2)
e^(x^2 - x - 2) = 2e^(x^2 - 2x - ln2)
e^(x^2 - x - 2) = e^(2x^2 - 4x - 2ln2)
e^(x^2 - x - 2) = e^(2x^2 - 4x) * e^(- 2ln2)
e^(x^2 - x - 2) = e^(2x^2 - 4x) * 0.25
e^(x^2 - x - 2) = e^0.25(2x^2 - 4x)
e^(x^2 - x - 2) = e^(0.5x^2 - x)
x^2 - x - 2 = 0.5x^2 - x
0.5x^2 = 2
x^2 = 4
x = 2
When x = 2 y = 4e^(2^2 - 2 - 2)
= 4e^0 = 4
So the point P is (2,4).
Find the equation of the normal to P:
To find the slope of the tangent at P we differentiate:-
The slope of the tangent = y' = 4* e^(x^2-x-2) * (2x - 1)
= (8x - 4)* e^(x^2 - x - 2)
When x = 2 this slope = (16-4) * e^0 = 12
So the slope of the normal = -1/12
y - y1 = m(x - x1)
y - 4 = -1/12(x - 2)
y = -1/12(x - 2) + 4
At point Q, y = 0 so
-1/12(x - 2) + 4 = 0
-1/12 x = -4 - 1/6 = -25/6
x = -25/6 * -12 = 50,
So point Q is at (50,0).
