TRIAL Find the domain and range of each of the following functions 1. \( h(x)=-2 x^{2}+12 x+5 \) 2. \( f(z)=|z-6| \) 3. \( g(x)=8 \) 4. \( h(x)=\frac{x}{\sqrt{x^{2}-9}} \) 5. \( g(t)=\sqrt{6+t-t^{2}} \)

Alright, I need to find the domain and range for each of these functions. Let's tackle them one by one. **1. \( h(x) = -2x^{2} + 12x + 5 \)** First, this is a quadratic function. The general form of a quadratic function is \( ax^{2} + bx + c \). For quadratic functions, the domain is all real numbers because there's no restriction on the input \( x \). So, the domain is \( (-\infty, \infty) \). Now, for the range. Since the coefficient of \( x^{2} \) is negative (\( -2 \)), the parabola opens downward. The vertex of the parabola will give the maximum value of the function. To find the vertex, I can use the formula \( x = -\frac{b}{2a} \). Here, \( a = -2 \) and \( b = 12 \), so: \[ x = -\frac{12}{2*(-2)} = -\frac{12}{-4} = 3 \] Now, plug \( x = 3 \) back into the function to find the y-coordinate of the vertex: \[ h(3) = -2*(3)^{2} + 12*3 + 5 = -18 + 36 + 5 = 23 \] So, the vertex is at \( (3, 23) \), and since the parabola opens downward, the maximum value of \( h(x) \) is 23. Therefore, the range is all real numbers less than or equal to 23, which is \( (-\infty, 23] \). **2. \( f(z) = |z - 6| \)** This is an absolute value function. The domain of an absolute value function is all real numbers because there's no restriction on the input \( z \). So, the domain is \( (-\infty, \infty) \). For the range, the absolute value function always outputs non-negative values. The smallest value it can take is 0, and it can grow infinitely large. Therefore, the range is \( [0, \infty) \). **3. \( g(x) = 8 \)** This is a constant function. The domain is all real numbers because there's no restriction on the input \( x \). So, the domain is \( (-\infty, \infty) \). For the range, since the function always outputs 8, regardless of the input, the range is just the single value 8, which is \( \{8\} \). **4. \( h(x) = \frac{x}{\sqrt{x^{2} - 9}} \)** First, I need to determine the domain. The denominator \( \sqrt{x^{2} - 9} \) must be defined and not equal to zero. So, \( x^{2} - 9 > 0 \), which means \( x^{2} > 9 \). Taking the square root of both sides, \( |x| > 3 \). Therefore, \( x < -3 \) or \( x > 3 \). So, the domain is \( (-\infty, -3) \cup (3, \infty) \). Now, for the range. The function \( h(x) = \frac{x}{\sqrt{x^{2} - 9}} \) can be rewritten as \( h(x) = \frac{x}{|x| \sqrt{1 - \frac{9}{x^{2}}}} \). As \( x \) approaches infinity or negative infinity, \( h(x) \) approaches 1 or -1, respectively. However, \( h(x) \) can take any real value except zero because the numerator and denominator are both non-zero and have the same sign, making \( h(x) \) always positive or always negative. Therefore, the range is \( (-\infty, 0) \cup (0, \infty) \). **5. \( g(t) = \sqrt{6 + t - t^{2}} \)** First, determine the domain. The expression inside the square root must be non-negative: \[ 6 + t - t^{2} \geq 0 \] Rearranging: \[ -t^{2} + t + 6 \geq 0 \] Multiply both sides by -1 (which reverses the inequality): \[ t^{2} - t - 6 \leq 0 \] Factor the quadratic: \[ (t - 3)(t + 2) \leq 0 \] The roots are \( t = 3 \) and \( t = -2 \). Testing intervals: - For \( t < -2 \): \( (t - 3)(t + 2) > 0 \) - For \( -2 < t < 3 \): \( (t - 3)(t + 2) < 0 \) - For \( t > 3 \): \( (t - 3)(t + 2) > 0 \) So, the domain is \( [-2, 3] \). For the range, since the function is a square root, the output is always non-negative. The maximum value occurs at the vertex of the quadratic inside the square root. The quadratic \( -t^{2} + t + 6 \) opens downward, so the maximum occurs at \( t = -\frac{b}{2a} = \frac{1}{2} \). Plugging \( t = \frac{1}{2} \) into the quadratic: \[ 6 + \frac{1}{2} - \left(\frac{1}{2}\right)^{2} = 6 + 0.5 - 0.25 = 6.25 \] So, the maximum value of the function is \( \sqrt{6.25} = 2.5 \). Therefore, the range is \( [0, 2.5] \).