Question

1. $l(x)=2f(x - 5)$

stretched by 2
right 5 units

1. $d(x)=f(\frac{1}{2}x)-4$
2. $n(x)=f(-x)+1$

Explanation:

First, identify key points of the parent function \(f(x)\) from the given graph:
Key points of \(f(x)\): \((-3, 8)\), \((0, 0)\), \((2, -4)\), \((3, 0)\), \((5, 8)\)

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For \(l(x)=2f(x-5)\):

Step1: Shift right 5 units

For each point \((x,y)\) of \(f(x)\), new \(x\)-coordinate: \(x+5\)
New points after shift: \((-3+5,8)=(2,8)\), \((0+5,0)=(5,0)\), \((2+5,-4)=(7,-4)\), \((3+5,0)=(8,0)\), \((5+5,8)=(10,8)\)

Step2: Vertically stretch by factor 2

New \(y\)-coordinate: \(2y\)
Final points: \((2, 16)\), \((5, 0)\), \((7, -8)\), \((8, 0)\), \((10, 16)\)

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For \(d(x)=f(\frac{1}{2}x)-4\):

Step1: Horizontally stretch by factor 2

For each point \((x,y)\) of \(f(x)\), new \(x\)-coordinate: \(2x\)
New points after stretch: \((-6,8)\), \((0,0)\), \((4,-4)\), \((6,0)\), \((10,8)\)

Step2: Shift down 4 units

New \(y\)-coordinate: \(y-4\)
Final points: \((-6, 4)\), \((0, -4)\), \((4, -8)\), \((6, -4)\), \((10, 4)\)

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For \(n(x)=f(-x)+1\):

Step1: Reflect over y-axis

For each point \((x,y)\) of \(f(x)\), new \(x\)-coordinate: \(-x\)
New points after reflection: \((3,8)\), \((0,0)\), \((-2,-4)\), \((-3,0)\), \((-5,8)\)

Step2: Shift up 1 unit

New \(y\)-coordinate: \(y+1\)
Final points: \((3, 9)\), \((0, 1)\), \((-2, -3)\), \((-3, 1)\), \((-5, 9)\)

Answer:

1. For \(l(x)=2f(x-5)\), plot the points \((2, 16)\), \((5, 0)\), \((7, -8)\), \((8, 0)\), \((10, 16)\) and connect them to match the shape of \(f(x)\).
2. For \(d(x)=f(\frac{1}{2}x)-4\), plot the points \((-6, 4)\), \((0, -4)\), \((4, -8)\), \((6, -4)\), \((10, 4)\) and connect them to match the shape of \(f(x)\).
3. For \(n(x)=f(-x)+1\), plot the points \((3, 9)\), \((0, 1)\), \((-2, -3)\), \((-3, 1)\), \((-5, 9)\) and connect them to match the shape of \(f(x)\).