Question

height of the object above the ground, in feet, x seconds after being launched. find and interpret the given function values and determine an appropriate domain for the function. answer attempt 1 out of 2 f(-2)=, meaning that seconds after the object was launched, the object was feet above the ground. this interpretation in the context of the problem. f(0.5)=, meaning that sec makes sense does not make sense lunched, the object was feet above the ground. this interpretation in the context of the problem. f(6)=, meaning that seconds after the object was launched, the object was feet above the ground. this interpretation in the context of the problem. based on the observations above, it is clear that an appropriate domain for the function is

Explanation:

1. Assume the function is \(y = f(x)=- 16x^{2}+10x + 50\) (since the original - function is not given, a common quadratic - function for projectile motion is assumed here).

• For \(f(-2)\):
• Explanation: Substitute \(x=-2\) into the function \(y = f(x)=-16x^{2}+10x + 50\).
• Step 1: Calculate the value of \(-16x^{2}\) when \(x = - 2\)
• \(-16x^{2}=-16\times(-2)^{2}=-16\times4=-64\).
• Step 2: Calculate the value of \(10x\) when \(x=-2\)
• \(10x = 10\times(-2)=-20\).
• Step 3: Calculate \(f(-2)\)
• \(f(-2)=-64-20 + 50=-34\).
• \(f(-2)=-34\), meaning that \(-2\) seconds after the object was launched, the object was \(-34\) feet above the ground. This interpretation does NOT make sense in the context of the problem because time \(x\) represents the time after the object is launched, and negative - time before the launch is not relevant in the physical context of the object's motion after being launched.
• For \(f(0.5)\):
• Explanation: Substitute \(x = 0.5\) into the function \(y = f(x)=-16x^{2}+10x + 50\).
• Step 1: Calculate the value of \(-16x^{2}\) when \(x = 0.5\)
• \(-16x^{2}=-16\times(0.5)^{2}=-16\times0.25=-4\).
• Step 2: Calculate the value of \(10x\) when \(x = 0.5\)
• \(10x=10\times0.5 = 5\).
• Step 3: Calculate \(f(0.5)\)
• \(f(0.5)=-4 + 5+50=51\).
• \(f(0.5)=51\), meaning that \(0.5\) seconds after the object was launched, the object was \(51\) feet above the ground. This interpretation makes sense in the context of the problem.
• For \(f(6)\):
• Explanation: Substitute \(x = 6\) into the function \(y = f(x)=-16x^{2}+10x + 50\).
• Step 1: Calculate the value of \(-16x^{2}\) when \(x = 6\)
• \(-16x^{2}=-16\times6^{2}=-16\times36=-576\).
• Step 2: Calculate the value of \(10x\) when \(x = 6\)
• \(10x=10\times6 = 60\).
• Step 3: Calculate \(f(6)\)
• \(f(6)=-576+60 + 50=-466\).
• \(f(6)=-466\), meaning that \(6\) seconds after the object was launched, the object was \(-466\) feet above the ground. This interpretation does NOT make sense in the context of the problem because the object cannot be below the ground level (negative height) in the context of the problem's physical situation (assuming the ground is at \(y = 0\)).

1. Determine the domain:

• In the physical context of the problem, the object is launched at \(x = 0\) and hits the ground when \(y=f(x)=0\).
• Set \(-16x^{2}+10x + 50 = 0\), divide through by \(-2\) to get \(8x^{2}-5x - 25=0\).
• Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\), here \(a = 8\), \(b=-5\), \(c=-25\).
• \(x=\frac{5\pm\sqrt{(-5)^{2}-4\times8\times(-25)}}{2\times8}=\frac{5\pm\sqrt{25 + 800}}{16}=\frac{5\pm\sqrt{825}}{16}=\frac{5\pm5\sqrt{33}}{16}\).
• We take the positive root \(x=\frac{5 + 5\sqrt{33}}{16}\approx\frac{5+5\times5.745}{16}=\frac{5+28.725}{16}\approx2.11\).
• The domain of the function is \(0\leq x\leq\frac{5 + 5\sqrt{33}}{16}\approx2.11\) (in the context of the physical problem, time starts at \(x = 0\) when the object is launched and ends when it hits the ground).

Answer:

\(f(-2)=-34\), meaning that \(-2\) seconds after the object was launched, the object was \(-34\) feet above the ground. This interpretation does NOT make sense in the context of the problem.
\(f(0.5)=51\), meaning that \(0.5\) seconds after the object was launched, the object was \(51\) feet above the ground. This interpretation makes sense in the context of the problem.
\(f(6)=-466\), meaning that \(6\) seconds after the object was launched, the object was \(-466\) feet above the ground. This interpretation does NOT make sense in the context of the problem.
The domain of the function is \(0\leq x\leq\frac{5 + 5\sqrt{33}}{16}\approx2.11\).