Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
1. Section 9.1
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Exercise 29 Page 427

Practice makes perfect
a We want to write the equation of the given parabola. To do so, let's recall the graphing form of a quadratic function.
y= a(x- h)^2+k In this expression, a, h, and k are either positive or negative constants. Let's start by identifying the vertex.

The vertex of this parabola has coordinates ( -1,-4). This means that we have h= -1 and k=- 4. We can use these values to partially write the equation of our function. y= a(x-( -1))^2+(- 4) ⇕ y= a(x+1)^2-4 We can see in the graph that the parabola opens upwards. Thus, a will be a positive number. To find its value, we will choose one point lying on the parabola that is not the vertex.

We can see above that the point has coordinates (1,0). Since this point is on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute 1 for x and 0 for y and simplify.
y=a(x+1)^2-4
0=a( 1+1)^2-4
â–Ľ
Solve for a
0=a(2)^2-4
0=a(4)-4
4=a(4)
1=a
a=1
We found that a= 1. Now we can complete the equation of the curve. y= 1(x+1)^2-4 ⇕ y=(x+1)^2-4
b We want to find the equation of the function represented by the table.
x - 4 - 3 - 2 - 1 0 1 2 3 4
y 12 5 0 - 3 - 4 - 3 0 5 12
Let's plot those points and connect them with a smooth curve to get the general idea about the type of the function.

As we can see, the relation represented by the table is a quadratic function. Thus, we want to write a quadratic equation represented by this function. To do this, we will use the points from the table to obtain the factored form of a quadratic equation. Factored form y=a(x+b)(x+c) In the factored form, - b and - c are the x-intercepts of the function. Those are the points where a graph crosses the y-axis, thus for those points y=0.

x - 4 - 3 - 2 - 1 0 1 2 3 4
y 12 5 0 - 3 - 4 - 3 0 5 12

As we can see, our x-intercepts are - 2 and 2. Therefore - b=- 2 and - c= 2, which means that b=2 and c=-2. Now we can partially complete the equation. y=a(x+b)(x+c) ⇕ y=a( x-2 ) ( x+2 ) We can see in the graph that the parabola opens upwards. Thus, a will be a positive number. To find its value, we will choose one point from the table that is not the x-intercept.

x - 4 - 3 - 2 - 1 0 1 2 3 4
y 12 5 0 - 3 - 4 - 3 0 5 12
Let's choose the point ( 0, - 4). Since this point is on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute 0 for x and - 4 for y and simplify.
y=a( x-2 ) ( x+2 )
- 4=a( 0-2 ) ( 0+2 )
â–Ľ
Solve for a
- 4=a(- 2)(2)
- 4=a(- 4)
1=a
a=1
We found that a= 1. Now we can complete the equation of the curve. y= 1( x-2 ) ( x+2 ) ⇕ y=( x-2 ) ( x+2 )