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. To find the value of a we can substitute 1 for x and 0 for y and simplify.
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 a 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. To find the value of a we can substitute 0 for x and - 4 for y and simplify.
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 )
c Comparing the graphs of the functions we made in Parts A and B, we notice that neither of them is bounded from above. As such there is no maximum value for either of them, as they both increase indefinitely.
