Directions: You must show your work.

1. [tex]y=(x-7)^2-3[/tex]

What is the vertex?
Transformation:
Does the function have a minimum or maximum?
What is the minimum or maximum value?
Fill out the table. The vertex goes in the middle.





Domain:

2. [tex]f(x)=-3(x+1)^2+5[/tex]

What is the vertex?
Transformation:
Does the function have a minimum or maximum?
What is the minimum or maximum value?
Fill out the table. The vertex goes in the middle.





Domain:

The vertex of the first function y = ( x − 7 ) 2 − 3 is ( 7 , − 3 ) , and it has a minimum value of − 3 .
The vertex of the second function f ( x ) = − 3 ( x + 1 ) 2 + 5 is ( − 1 , 5 ) , and it has a maximum value of 5 .
Both functions have a domain of all real numbers.
The transformations and tables of values are as described above.

See full solution above. ​
Explanation

Problem Analysis We are given two quadratic functions and asked to analyze their properties, including finding the vertex, describing the transformations, determining if they have a minimum or maximum value, finding that value, creating a table of values, and stating the domain.

Finding the Vertex of the First Function For the first function, y = ( x − 7 ) 2 − 3 , we can identify the vertex directly from the vertex form of a quadratic equation, which is y = a ( x − h ) 2 + k , where ( h , k ) is the vertex. In this case, h = 7 and k = − 3 , so the vertex is ( 7 , − 3 ) .

Describing the Transformation of the First Function The transformation of the first function can be described as a horizontal shift of 7 units to the right and a vertical shift of 3 units down, compared to the parent function y = x 2 .

Determining Minimum or Maximum for the First Function Since the coefficient of the ( x − 7 ) 2 term is positive (1), the parabola opens upwards, and the function has a minimum value.

Finding the Minimum Value of the First Function The minimum value of the first function is the y-coordinate of the vertex, which is − 3 .

Creating a Table of Values for the First Function Now, let's create a table of values for the first function. We'll center the table around the vertex, so we'll choose x-values of 5, 6, 7, 8, and 9. We can calculate the corresponding y-values as follows:


For x = 5 , y = ( 5 − 7 ) 2 − 3 = ( − 2 ) 2 − 3 = 4 − 3 = 1 .
For x = 6 , y = ( 6 − 7 ) 2 − 3 = ( − 1 ) 2 − 3 = 1 − 3 = − 2 .
For x = 7 , y = ( 7 − 7 ) 2 − 3 = ( 0 ) 2 − 3 = 0 − 3 = − 3 .
For x = 8 , y = ( 8 − 7 ) 2 − 3 = ( 1 ) 2 − 3 = 1 − 3 = − 2 .
For x = 9 , y = ( 9 − 7 ) 2 − 3 = ( 2 ) 2 − 3 = 4 − 3 = 1 .
So, the table is:



x

y



5

1


6

-2


7

-3


8

-2


9

1



Stating the Domain of the First Function The domain of the first function is all real numbers, since it's a quadratic function.

Finding the Vertex of the Second Function For the second function, f ( x ) = − 3 ( x + 1 ) 2 + 5 , we can identify the vertex directly from the vertex form of a quadratic equation, which is f ( x ) = a ( x − h ) 2 + k , where ( h , k ) is the vertex. In this case, h = − 1 and k = 5 , so the vertex is ( − 1 , 5 ) .

Describing the Transformation of the Second Function The transformation of the second function can be described as a horizontal shift of 1 unit to the left, a vertical shift of 5 units up, a reflection over the x-axis (due to the negative coefficient), and a vertical stretch by a factor of 3.

Determining Minimum or Maximum for the Second Function Since the coefficient of the ( x + 1 ) 2 term is negative (-3), the parabola opens downwards, and the function has a maximum value.

Finding the Maximum Value of the Second Function The maximum value of the second function is the y-coordinate of the vertex, which is 5 .

Creating a Table of Values for the Second Function Now, let's create a table of values for the second function. We'll center the table around the vertex, so we'll choose x-values of -3, -2, -1, 0, and 1. We can calculate the corresponding y-values as follows:


For x = − 3 , f ( − 3 ) = − 3 ( − 3 + 1 ) 2 + 5 = − 3 ( − 2 ) 2 + 5 = − 3 ( 4 ) + 5 = − 12 + 5 = − 7 .
For x = − 2 , f ( − 2 ) = − 3 ( − 2 + 1 ) 2 + 5 = − 3 ( − 1 ) 2 + 5 = − 3 ( 1 ) + 5 = − 3 + 5 = 2 .
For x = − 1 , f ( − 1 ) = − 3 ( − 1 + 1 ) 2 + 5 = − 3 ( 0 ) 2 + 5 = − 3 ( 0 ) + 5 = 0 + 5 = 5 .
For x = 0 , f ( 0 ) = − 3 ( 0 + 1 ) 2 + 5 = − 3 ( 1 ) 2 + 5 = − 3 ( 1 ) + 5 = − 3 + 5 = 2 .
For x = 1 , f ( 1 ) = − 3 ( 1 + 1 ) 2 + 5 = − 3 ( 2 ) 2 + 5 = − 3 ( 4 ) + 5 = − 12 + 5 = − 7 .
So, the table is:



x

y



-3

-7


-2

2


-1

5


0

2


1

-7



Stating the Domain of the Second Function The domain of the second function is all real numbers, since it's a quadratic function.

Examples
Understanding quadratic functions is crucial in various real-world applications. For instance, engineers use quadratic equations to design parabolic mirrors and reflectors, which are used in telescopes and satellite dishes. The vertex of the parabola helps determine the optimal focal point for these devices, maximizing their efficiency. Similarly, in projectile motion, the height of an object thrown into the air can be modeled by a quadratic function, where the vertex represents the maximum height reached by the object.

Answered by GinnyAnswer | 2025-07-08

The vertex of the first function is (7, -3) with a minimum value of -3. The vertex of the second function is (-1, 5) with a maximum value of 5. Both functions have a domain of all real numbers.
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Answered by Anonymous | 2025-07-23