Question

the function $f(x) = -x^2 - 2x + 15$ is shown on the graph. what are the domain and range of the function?

• the domain is all real numbers. the range is ${y|y < 16}$.
• the domain is all real numbers. the range is ${y|y \leq 16}$.
• the domain is ${x|-5 < x < 3}$. the range is ${y|y < 16}$.
• the domain is ${x|-5 \leq x \leq 3}$. the range is ${y|y \leq 16}$.

Explanation:

Step1: Analyze the domain of a quadratic function

A quadratic function \( f(x) = -x^2 - 2x + 15 \) is a polynomial function. The domain of any polynomial function is all real numbers because there are no restrictions on the values of \( x \) that can be plugged into the function (no division by zero, no square roots of negative numbers, etc.). So the domain is all real numbers, which eliminates the options where the domain is restricted to \( - 5 < x < 3 \) or \( - 5\leqslant x\leqslant3 \).

Step2: Analyze the range of the quadratic function

The function \( f(x)=-x^{2}-2x + 15 \) is a quadratic function in the form \( y = ax^{2}+bx + c \), where \( a=- 1 \), \( b=-2 \), \( c = 15 \). Since \( a=-1<0 \), the parabola opens downward. The vertex of a parabola \( y=ax^{2}+bx + c \) has its \( x \)-coordinate given by \( x=-\frac{b}{2a} \). Substituting \( a=-1 \) and \( b = - 2 \), we get \( x=-\frac{-2}{2\times(-1)}=-\frac{-2}{-2}=-1 \). To find the \( y \)-coordinate of the vertex, substitute \( x = - 1 \) into the function: \( f(-1)=-(-1)^{2}-2\times(-1)+15=-1 + 2+15 = 16 \). Since the parabola opens downward, the maximum value of the function is \( 16 \), and the function can take any value less than or equal to \( 16 \). So the range is \( \{y|y\leqslant16\} \).

Answer:

The domain is all real numbers. The range is \(\{y|y\leqslant16\}\) (the second option: The domain is all real numbers. The range is \(\{y|y\leqslant16\}\))