Math  /  Algebra

Question18. Tyler and Kym are competing in a model airplane competition. The positions of the planes are determined in relation to the top of a 10 -metre pole: planes to the left and below the top of the pole are given negative xx - and yy-values and planes to the right and above the top of the pole are given positive xx - and yy-values. The path of Tyler's plane is modeled by f(x)=4x36x222x+8f(x)=4 x^{3}-6 x^{2}-22 x+8 The path of Kym's plane is modeled by g(x)=3x33x29x7g(x)=3 x^{3}-3 x^{2}-9 x-7 a) Set up an inequality to determine when Tyler's plane is higher that Kym's plane. (1 mark) f(x)>g(x)=4x36x222x+8>3x33x29x7f(x)>g(x)=4 x^{3}-6 x^{2}-22 x+8>3 x^{3}-3 x^{2}-9 x-7 b) What equation would have to be entered into the graphing calculator to solve the inequality in part a? ( 1 mark) (4x36x222x+8)(3x33x29x7)>04x36x22x+83x2+3x2+9x+7>0x33x213x+15>0\begin{array}{c} \left(4 x^{3}-6 x^{2}-22 x+8\right)-\left(3 x^{3}-3 x^{2}-9 x-7\right)>0 \\ 4 x^{3}-6 x^{2}-2 x+8-3 x^{2}+3 x^{2}+9 x+7>0 \\ x^{3}-3 x^{2}-13 x+15>0 \end{array} c) Write the inequality in factored form and solve by graphing to determine when is Tyler's plane higher than Kym's?. (5 marks) factored pomm: x33x213x+15=0x^{3}-3 x^{2}-13 x+15=0.

Studdy Solution
Interpret the solution from the graph:
The inequality (x1)(x3)(x+5)>0 (x - 1)(x - 3)(x + 5) > 0 is satisfied in the intervals where the product is positive. Typically, this occurs in intervals such as (,5)(1,3) (-\infty, -5) \cup (1, 3) .
Thus, Tyler's plane is higher than Kym's plane in these intervals.
The intervals where Tyler's plane is higher than Kym's are:
(,5)(1,3) (-\infty, -5) \cup (1, 3)

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