y =32- ax2 ;
y(28)=y(-28)=0;
32- a*282=0,
"a=\\frac{32}{28^2};"
"a=\\frac{2}{49}" ;
"y=32-\\frac{2}{49}x^2."
Next, one should find x such that "y \\geq 22"
"32-\\frac{2}{49}x^2 \\geq 22,"
"\\frac{2}{49}x^2 \\leq 10,"
"x^2 \\leq \\frac{490}{2},"
"-7 \\sqrt{5} \\leq x \\leq 7 \\sqrt{5},"
-15.66<x<15.66
If the vertex of parabola is (0,32), then a person should be located between -15.66 and 15.66, that is, approximately 15.66 inches width to the left of the vertex of the parabola and approximately 15.66 inches width to the right of the vertex of the parabola where the parabola is higher than the height of 22 inches.
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Dear Samuel kassapa, more details were described in a solution of the question.
Please am not clear of the solution to the question #124581 Please I need more clarity.thanks
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