Use a graphing calculator to sketch the graph of the quadratic equation, and then state the domain and range. [tex]y=x^2+2 x+2[/tex]
A. D: all real numbers
R: [tex](y \leq 2)[/tex]
B. D : all real numbers
R :[tex](y \leq 1)[/tex]
C. D: [tex](x \geq 1)[/tex]
R: [tex](y \geq 2)[/tex]
D. D : all real numbers
R: [tex](y \geq 1)[/tex]
Please select the best answer from the choices provided
Answer (2)
The domain of the quadratic equation y = x 2 + 2 x + 2 is all real numbers, while the range is y ≥ 1 . Therefore, the best answer choice is D. This means the equation can take any real number for x , and the smallest value for y is 1.
;
The domain of the quadratic equation y = x 2 + 2 x + 2 is all real numbers.
Complete the square to find the vertex form: y = ( x + 1 ) 2 + 1 . The vertex is ( − 1 , 1 ) .
Since the parabola opens upwards, the range is y ≥ 1 .
The correct answer is D, with domain as all real numbers and range as y ≥ 1 .
Explanation
Understanding Domain and Range We are given the quadratic equation y = x 2 + 2 x + 2 . Our goal is to determine its domain and range. The domain represents all possible x values that can be input into the equation, while the range represents all possible y values that the equation can output.
Determining the Domain Since y = x 2 + 2 x + 2 is a polynomial, we can input any real number for x . There are no restrictions, such as division by zero or taking the square root of a negative number. Therefore, the domain is all real numbers.
Finding the Range To find the range, we need to determine the minimum or maximum value of the quadratic function. We can do this by completing the square to rewrite the equation in vertex form, which is y = a ( x − h ) 2 + k , where ( h , k ) is the vertex of the parabola. The vertex represents the minimum or maximum point of the parabola.
Completing the Square Completing the square for y = x 2 + 2 x + 2 :
y = ( x 2 + 2 x ) + 2
To complete the square, we need to add and subtract ( 2/2 ) 2 = 1 inside the parenthesis:
y = ( x 2 + 2 x + 1 − 1 ) + 2
y = ( x 2 + 2 x + 1 ) − 1 + 2
y = ( x + 1 ) 2 + 1
So, the vertex form of the equation is y = ( x + 1 ) 2 + 1 .
Identifying the Vertex From the vertex form y = ( x + 1 ) 2 + 1 , we can see that the vertex of the parabola is ( − 1 , 1 ) . Since the coefficient of the x 2 term is positive (1), the parabola opens upwards. This means that the vertex represents the minimum point of the parabola.
Determining the Range Since the parabola opens upwards and the vertex is at ( − 1 , 1 ) , the minimum value of y is 1. Therefore, the range of the quadratic function is all y values greater than or equal to 1, which can be written as y ≥ 1 .
Selecting the Correct Option The domain is all real numbers, and the range is y ≥ 1 . Comparing this with the given options, we see that option D matches our result.
Final Answer Therefore, the correct answer is D: Domain: all real numbers, Range: y ≥ 1 .
Examples
Quadratic equations are used in various real-world applications, such as modeling the trajectory of a projectile, designing parabolic mirrors and reflectors, and optimizing processes in engineering and economics. For example, if you throw a ball, its path can be modeled by a quadratic equation, and understanding the domain and range helps determine the possible distances the ball can travel and the maximum height it can reach. Similarly, engineers use quadratic equations to design satellite dishes that focus signals onto a receiver, ensuring optimal performance within a specific range of angles.
