Answer :
Let's analyze the function [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] and identify its characteristics step-by-step:
1. Domain:
The domain of a quadratic function, or indeed any polynomial function, is all real numbers. Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers.
2. Range:
The function [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] is in the vertex form of a quadratic function. The vertex form is [tex]\( (x-h)^2 + k \)[/tex], where [tex]\( (h, k) \)[/tex] is the vertex of the parabola. For this function, the vertex [tex]\( (-1, 2) \)[/tex] is the minimum point because the parabola opens upwards (the coefficient of the squared term is positive). Thus, the range is all real numbers greater than or equal to 2.
3. Y-Intercept:
The y-intercept is found by setting [tex]\( x = 0 \)[/tex] in the function. Thus,
[tex]\[ f(0) = (0 + 1)^2 + 2 = 1 + 2 = 3. \][/tex]
Hence, the y-intercept is 3.
4. Transformation:
To understand the transformation, we note that [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] can be seen as a transformation of [tex]\( y = x^2 \)[/tex]:
- The term [tex]\( x + 1 \)[/tex] indicates a horizontal shift 1 unit to the left.
- The [tex]\( +2 \)[/tex] indicates a vertical shift 2 units up.
Therefore, the graph is 1 unit to the left and 2 units up from the graph of [tex]\( y = x^2 \)[/tex].
5. X-Intercepts:
The x-intercepts are found by setting [tex]\( f(x) = 0 \)[/tex] and solving for [tex]\( x \)[/tex]:
[tex]\[ (x + 1)^2 + 2 = 0. \][/tex]
Simplifying gives:
[tex]\[ (x + 1)^2 = -2. \][/tex]
Since the square of a real number cannot be negative, this equation has no real solutions. Therefore, the graph does not have any x-intercepts.
Based on this analysis, the characteristics of the graph of [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] are:
- The domain is all real numbers.
- The range is all real numbers greater than or equal to 2.
- The y-intercept is 3.
- The graph of the function is 1 unit to the left and 2 units up from the graph of [tex]\( y = x^2 \)[/tex].
- The graph has no x-intercepts.
Let's check the given statements:
1. "The domain is all real numbers." → True.
2. "The range is all real numbers greater than or equal to 1." → False (It should be 2, not 1).
3. "The [tex]\( y \)[/tex]-intercept is 3." → True.
4. "The graph of the function is 1 unit up and 2 units to the left from the graph of [tex]\( y = x^2 \)[/tex]." → False (It is 1 unit to the left and 2 units up, not 1 unit up and 2 units to the left).
5. "The graph has two [tex]\( x \)[/tex]-intercepts." → False (There are no [tex]\( x \)[/tex]-intercepts).
Thus, the correct checked characteristics are:
- The domain is all real numbers.
- The [tex]\( y \)[/tex]-intercept is 3.
1. Domain:
The domain of a quadratic function, or indeed any polynomial function, is all real numbers. Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers.
2. Range:
The function [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] is in the vertex form of a quadratic function. The vertex form is [tex]\( (x-h)^2 + k \)[/tex], where [tex]\( (h, k) \)[/tex] is the vertex of the parabola. For this function, the vertex [tex]\( (-1, 2) \)[/tex] is the minimum point because the parabola opens upwards (the coefficient of the squared term is positive). Thus, the range is all real numbers greater than or equal to 2.
3. Y-Intercept:
The y-intercept is found by setting [tex]\( x = 0 \)[/tex] in the function. Thus,
[tex]\[ f(0) = (0 + 1)^2 + 2 = 1 + 2 = 3. \][/tex]
Hence, the y-intercept is 3.
4. Transformation:
To understand the transformation, we note that [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] can be seen as a transformation of [tex]\( y = x^2 \)[/tex]:
- The term [tex]\( x + 1 \)[/tex] indicates a horizontal shift 1 unit to the left.
- The [tex]\( +2 \)[/tex] indicates a vertical shift 2 units up.
Therefore, the graph is 1 unit to the left and 2 units up from the graph of [tex]\( y = x^2 \)[/tex].
5. X-Intercepts:
The x-intercepts are found by setting [tex]\( f(x) = 0 \)[/tex] and solving for [tex]\( x \)[/tex]:
[tex]\[ (x + 1)^2 + 2 = 0. \][/tex]
Simplifying gives:
[tex]\[ (x + 1)^2 = -2. \][/tex]
Since the square of a real number cannot be negative, this equation has no real solutions. Therefore, the graph does not have any x-intercepts.
Based on this analysis, the characteristics of the graph of [tex]\( f(x) = (x+1)^2 + 2 \)[/tex] are:
- The domain is all real numbers.
- The range is all real numbers greater than or equal to 2.
- The y-intercept is 3.
- The graph of the function is 1 unit to the left and 2 units up from the graph of [tex]\( y = x^2 \)[/tex].
- The graph has no x-intercepts.
Let's check the given statements:
1. "The domain is all real numbers." → True.
2. "The range is all real numbers greater than or equal to 1." → False (It should be 2, not 1).
3. "The [tex]\( y \)[/tex]-intercept is 3." → True.
4. "The graph of the function is 1 unit up and 2 units to the left from the graph of [tex]\( y = x^2 \)[/tex]." → False (It is 1 unit to the left and 2 units up, not 1 unit up and 2 units to the left).
5. "The graph has two [tex]\( x \)[/tex]-intercepts." → False (There are no [tex]\( x \)[/tex]-intercepts).
Thus, the correct checked characteristics are:
- The domain is all real numbers.
- The [tex]\( y \)[/tex]-intercept is 3.
