Answer:
Step-by-step explanation:
From the question we are told that
Altitude of Height 1.4miles
4miles west
a) From the question
[tex]f(x)=ax3+bx2+cx+d[/tex] on interval (-4,0)
The descent can be expressed in [tex]\triangle _x =-4,0[/tex]
therefore initial points is (-4 , 1.4).
final co-ordinate is given by (0,0) as its landing
Mathematically solving in above cubic equation
[tex]0 = 0 + 0 + 0 + d \\[/tex]
[tex]d = 0[/tex]
[tex]1.4 = -64a + 16b -4c + d[/tex]
[tex]1 = -45.7a + 11.4b-2.9c+d[/tex]
Generally in this case Slope of tangent at point of descent is zero.assuming flight was initially straight before descent
[tex]f'(x) = 3ax^2 + 2bx+c[/tex]
[tex]f'(-4) = 48a -8b + c = 0[/tex]
[tex]f'(0) = c = 0[/tex]
Mathematically solving above equations we have
[tex]a = \frac{1}{40} \ and \ b = \frac{3}{20}[/tex]
Therefore having found a and b the Cubic equation is given as
[tex]f(x) = \frac{1}{40}x^3 + \frac{3}{20}x^2[/tex]
b)descent at its greatest rate
[tex]f(x)=\frac{1.4}{40} x^3 + \frac{129}{20} x^2[/tex]
[tex]f(x)=\frac{1.4}{40} x^3 + \frac{129}{20} x^2[/tex]
