Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 8.2
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Exercise 106 Page 406

Practice makes perfect
a To rewrite the given quadratic function in graphing form, we will start by calculating ( b2)^2.
f(x)=x^2+6x+15We can see that b=6. Let's use this to calculate ( b2 )^2.
( b/2 )^2
( 6/2 )^2
3^2
Let's now factor the function by completing the square. We will first add ( b2 )^2= 3^2. Note that we also have to subtract 3^2 to leave the function unchanged.
f(x)=x^2+6x+15
f(x)=x^2+6x+ 3^2+15- 3^2
f(x)=(x+3)^2+15-3^2
f(x)=(x+3)^2+15-9
f(x)=(x+3)^2+6
Now we know that the graphing form is f(x)=(x+3)^2+6, which can be also written as f(x)=(x-( - 3))^2+6. The vertex of the parabola is ( - 3,6).
b To rewrite the given quadratic function in graphing form, we will start by calculating ( b2)^2.
f(x)=x^2-4x+9We can see that b=- 4. Let's use this to calculate ( b2 )^2.
( b/2 )^2
( - 4/2 )^2
â–Ľ
Simplify
( - 4/2 )^2
(- 2)^2
2^2
Let's now factor the function by completing the square. We will first add ( b2 )^2= 2^2. Note that we also have to subtract 2^2 to leave the function unchanged.
f(x)=x^2-4x+9
f(x)=x^2-4x+ 2^2+9- 2^2
f(x)=(x-2)^2+9-2^2
f(x)=(x-2)^2+9-4
f(x)=(x-2)^2+5
Now we know that the vertex form is f(x)=(x- 2)^2+5 and the vertex of the parabola is ( 2,5).
c To rewrite the given quadratic function in graphing form, we will start by calculating ( b2)^2.
f(x)=x^2+8xWe can see that b=8. Let's use this to calculate ( b2 )^2.
( b/2 )^2
( 8/2 )^2
4^2
Let's now factor the function by completing the square. We will first add ( b2 )^2= 4^2. Note that we also have to subtract 4^2 to leave the function unchanged.
f(x)=x^2+8x
f(x)=x^2+8x+ 4^2- 4^2
f(x)=(x+4)^2-4^2
f(x)=(x+4)^2-16
Now we know that the graphing form is f(x)=(x+4)^2-16, which can be also written as f(x)=(x-( - 4))^2+(- 16). The vertex of the parabola is ( - 4,- 16).
d To rewrite the given quadratic function in graphing form, we will start by calculating ( b2)^2.
f(x)=x^2+5x-2We can see that b=5. Let's use this to calculate ( b2 )^2.
( b/2 )^2
( 5/2 )^2
Let's now factor the function by completing the square. We will first add ( b2 )^2= ( 52)^2. Note that we also have to subtract ( 52)^2 to leave the function unchanged.
f(x)=x^2+5x-2
f(x)=x^2+5x+ (5/2)^2-2- (5/2)^2
f(x)=(x+5/2)^2-2-(5/2)^2
â–Ľ
Simplify right-hand side
f(x)=(x+5/2)^2-2-5^2/2^2
f(x)=(x+5/2)^2-2-25/4
f(x)=(x+5/2)^2-8/4-25/4
f(x)=(x+5/2)^2-33/4
Now we know that the graphing form is (x+ 52)^2- 334, which can be also written as f(x)=(x-( - 52))^2+(- 334). The vertex of the parabola is ( - 52,- 334).