Question
The quadratic function f(x)=x2+2x−15
is shown in the graph. Determine the domain and range of this function.
(1 point)
Responses
domain R
; range R
domain the real numbers ; range the real numbers
domain R
; range y≤−16
domain negative infinity less-than-or-equal-to x less-than-or-equal-to infinity ; range y is less than or equal to negative 16
domain R
; range y≥−16
domain the real numbers ; range y is greater than or equal to negative 16
domain R
; range y≥16
is shown in the graph. Determine the domain and range of this function.
(1 point)
Responses
domain R
; range R
domain the real numbers ; range the real numbers
domain R
; range y≤−16
domain negative infinity less-than-or-equal-to x less-than-or-equal-to infinity ; range y is less than or equal to negative 16
domain R
; range y≥−16
domain the real numbers ; range y is greater than or equal to negative 16
domain R
; range y≥16
Answers
GPT-4o mini
To determine the domain and range of the quadratic function \( f(x) = x^2 + 2x - 15 \), let's analyze its features.
1. **Domain**: The domain of a quadratic function is always all real numbers unless restricted by some other condition (like a logarithmic or square root function). Therefore, for \( f(x) \), the domain is:
\[
\text{Domain: } \mathbb{R} \text{ (all real numbers)}
\]
2. **Range**: To find the range, we need to determine the vertex of the quadratic, as it opens upward (the coefficient of \( x^2 \) is positive). The vertex can be found using the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \):
\[
a = 1, \quad b = 2 \implies x = -\frac{2}{2 \cdot 1} = -1
\]
Next, we substitute \( x = -1 \) back into the function to find the y-coordinate of the vertex:
\[
f(-1) = (-1)^2 + 2(-1) - 15 = 1 - 2 - 15 = -16
\]
Since the parabola opens upward, the minimum value of \( f(x) \) occurs at the vertex and is \(-16\). Thus, the range of \( f(x) \) is:
\[
\text{Range: } y \geq -16
\]
Considering the options provided, the correct answer is:
- domain \( \mathbb{R} \); range \( y \geq -16 \)
1. **Domain**: The domain of a quadratic function is always all real numbers unless restricted by some other condition (like a logarithmic or square root function). Therefore, for \( f(x) \), the domain is:
\[
\text{Domain: } \mathbb{R} \text{ (all real numbers)}
\]
2. **Range**: To find the range, we need to determine the vertex of the quadratic, as it opens upward (the coefficient of \( x^2 \) is positive). The vertex can be found using the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \):
\[
a = 1, \quad b = 2 \implies x = -\frac{2}{2 \cdot 1} = -1
\]
Next, we substitute \( x = -1 \) back into the function to find the y-coordinate of the vertex:
\[
f(-1) = (-1)^2 + 2(-1) - 15 = 1 - 2 - 15 = -16
\]
Since the parabola opens upward, the minimum value of \( f(x) \) occurs at the vertex and is \(-16\). Thus, the range of \( f(x) \) is:
\[
\text{Range: } y \geq -16
\]
Considering the options provided, the correct answer is:
- domain \( \mathbb{R} \); range \( y \geq -16 \)