\( 1 \cdot 2 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 1 \cdot 3 \cdot 1 \cdot 4 \cdot 1 \cdot 5 \cdot 1 \cdot 6 \cdot 1 \cdot 7 \cdot 1 \cdot 8 \cdot 1 \cdot 9 \cdot 1 \cdot 10 \cdot 1 \cdot 11 \) Sketch the graph of \( f(x)=3 x-x^{2} \) and find a. the domain and range b. \( f(q) \) c. \( f\left(x^{2}\right) \) d. \( \frac{f(2+h) \cdot f(2)}{h}, h \neq 0 \).

Let's analyze the function \( f(x) = 3x - x^2 \). ### a. Domain and Range **Domain:** The function \( f(x) = 3x - x^2 \) is a polynomial function, and polynomial functions are defined for all real numbers. Therefore, the domain is: \[ \text{Domain: } (-\infty, \infty) \] **Range:** To find the range, we first determine the vertex of the parabola represented by the function. The function can be rewritten in standard form as: \[ f(x) = -x^2 + 3x \] This is a downward-opening parabola. The vertex \( x \) coordinate can be found using the formula \( x = -\frac{b}{2a} \), where \( a = -1 \) and \( b = 3 \): \[ x = -\frac{3}{2 \cdot -1} = \frac{3}{2} \] Now, we can find the maximum value of \( f(x) \) by substituting \( x = \frac{3}{2} \) back into the function: \[ f\left(\frac{3}{2}\right) = 3\left(\frac{3}{2}\right) - \left(\frac{3}{2}\right)^2 = \frac{9}{2} - \frac{9}{4} = \frac{18}{4} - \frac{9}{4} = \frac{9}{4} \] Thus, the maximum value of \( f(x) \) is \( \frac{9}{4} \). Since the parabola opens downwards, the range of \( f(x) \) is: \[ \text{Range: } (-\infty, \frac{9}{4}] \] ### b. \( f(q) \) To find \( f(q) \), we simply substitute \( q \) into the function: \[ f(q) = 3q - q^2 \] ### c. \( f(x^2) \) To find \( f(x^2) \), we substitute \( x^2 \) into the function: \[ f(x^2) = 3(x^2) - (x^2)^2 = 3x^2 - x^4 \] ### d. \( \frac{f(2+h) \cdot f(2)}{h}, h \neq 0 \) First, we need to calculate \( f(2) \): \[ f(2) = 3(2) - (2)^2 = 6 - 4 = 2 \] Next, we calculate \( f(2+h) \): \[ f(2+h) = 3(2+h) - (2+h)^2 = 3(2+h) - (4 + 4h + h^2) = 6 + 3h - 4 - 4h - h^2 = 2 - h - h^2 \] Now, we can substitute \( f(2+h) \) and \( f(2) \) into the expression: \[ \frac{f(2+h) \cdot f(2)}{h} = \frac{(2 - h - h^2) \cdot 2}{h} = \frac{2(2 - h - h^2)}{h} = \frac{4 - 2h - 2h^2}{h} = \frac{4}{h} - 2 - 2h \] Thus, the final expression is: \[ \frac{f(2+h) \cdot f(2)}{h} = \frac{4}{h} - 2 - 2h \] ### Summary of Results - **Domain:** \( (-\infty, \infty) \) - **Range:** \( (-\infty, \frac{9}{4}] \) - **\( f(q) \):** \( 3q - q^2 \) - **\( f(x^2) \):** \( 3x^2 - x^4 \) - **\( \frac{f(2+h) \cdot f(2)}{h} \):** \( \frac{4}{h} - 2 - 2h \)