1. Find the vertex form of each of the following quadratic functions. Find also y -intercept, axis of symmetry, vertex, and range of each of the functions.
(a) y=2x2+4x+3(b) y=3x2−6x+2(c) y=12x2+x−4(d) y=−2x2+2x+3(e) y=−3x2−12x−7(f) y=−12x2−3x−4
(a)y=2x2+4x+3
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y=2x2+4x+3 y=2(x2+2x+1)+1 y=2(x+1)2+1 When x=0, y=3 ∴ y−intercept (0, 3). Comparing with a(x−h)2+k, a=2, h=−1, k =1 ∴ axis of symmetry:x=h⇒x=−1 vertex=(h,k)=(−1,1) range={y | y ≥ k}={y | y ≥ 1}
(b)y=3x2−6x+2
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(c)y=12x2+x−4
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y=12x2+x−4 y=12x2+22x+12−92 y=12(x2+2x+1)−92 y=12(x+1)2−92 When x=0, y=−4 ∴ y−intercept (0, −4). Comparing with a(x−h)2+k, a=12, h=−1, k =−92 ∴ axis of symmetry: x=h⇒x=−1 vertex=(h,k)=(−1,−92) range={y | y ≥ k}={y | y ≥ −92}
(d) y=−2x2+2x+3
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y=−2x2+2x+3 y=−2(x2−x)+3 y=−2(x2−2(12)x+14)+3+12 y=−2(x−12)2+72 When x=0, y=3 ∴ y−intercept (0, 3). Comparing with a(x−h)2+k, a=−2, h=12, k =72 ∴ axis of symmetry: x=h⇒x=12 vertex=(h,k)=(12,72) range={y | y ≤ k}={y | y ≤ 72}
(e)y=−3x2−12x−7
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y=−3x2−12x−7y=−3(x2+4x)−7y=−3(x2+4x+4)−7+12y=−3(x+2)2+5Whenx=0,y=−7∴y−intercept(0,−7).Comparing with a(x−h)2+k,a=−3,h=−2,k=5∴axis of symmetry: x=h⇒x=−2vertex=(h,k)=(−2,5)range={y|y≤k}={y|y≤5}
(f)y=−12x2−3x−4
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y=−12x2−3x−4y=−12(x2+6x)−4y=−12(x2+2(3)x+9)−4+92y=−12(x+3)2+12Whenx=0,y=−4∴y−intercept(0,−4).Comparing with a(x−h)2+k,a=−12,h=−3,k=12∴axis of symmetry: x=h⇒x=−3vertex=(h,k)=(−3,12)range={y|y≤k}={y|y≤12}
2. Find two positive numbers whose sum is 20 and whose product is as large as possible.
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Let the two positive numbers be x and y.
By the problem,
x+y=20
∴ y=20−x
Let the product of the numbers be p.
Hence,
p=xy=x(20−x)=20x−x2=−(x2−20x)=−(x2−20x+100)+100=100−(x−10)2∴ p≤100
Thus, maximum value of p is 100 when (x−10)2=0.
∴ x=10.
When x=10,y=20−x=20−10=10.
Therefore, the twopositive numbers are x=10 and y=10.
By the problem,
x+y=20
∴ y=20−x
Let the product of the numbers be p.
Hence,
p=xy=x(20−x)=20x−x2=−(x2−20x)=−(x2−20x+100)+100=100−(x−10)2∴ p≤100
Thus, maximum value of p is 100 when (x−10)2=0.
∴ x=10.
When x=10,y=20−x=20−10=10.
Therefore, the twopositive numbers are x=10 and y=10.
3. What is the largest area possible for a rectangle whose perimeter is 16 cm?
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Let the lenght and breadth of the rectangle be x and y respectively.
By the problem,
2x+2y=16
∴ y=8−x
Let the area of the rectangle be A.
Hence,
A=xy=x(8−x)=8x−x2=−(x2−8x)=−(x2−2⋅4⋅x+16)+16=16−(x−4)2∴ A≤16
Thus, the maximum area (A) of the rectangle is 16 cm².
By the problem,
2x+2y=16
∴ y=8−x
Let the area of the rectangle be A.
Hence,
A=xy=x(8−x)=8x−x2=−(x2−8x)=−(x2−2⋅4⋅x+16)+16=16−(x−4)2∴ A≤16
Thus, the maximum area (A) of the rectangle is 16 cm².
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