4) \( h(x)=-x^{2} \) 4) \( j(x)=x^{2}+5 x-6 \) 6) \( K(x)=2 x^{2}+5 x-6 \) \( k(x)=\frac{1}{2} x^{2}-\frac{2}{x} x \)

To solve these study problems, we need to understand the functions and possibly find their properties such as domain, range, zeros, or any other specific information requested. Let's analyze each function: 1. \( h(x) = -x^2 \) - This is a quadratic function with a negative coefficient for \( x^2 \), which means it opens downwards and has a maximum point. - The vertex of the parabola is at the origin (0,0) because the function is symmetric about the y-axis. - The domain is all real numbers, \( (-\infty, \infty) \). - The range is all non-positive real numbers, \( (-\infty, 0] \). 2. \( j(x) = x^2 + 5x - 6 \) - This is also a quadratic function, which opens upwards because the coefficient for \( x^2 \) is positive. - To find the vertex, we can use the formula \( x = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 5 \). - The vertex is at \( x = -\frac{5}{2} \). - The domain is all real numbers, \( (-\infty, \infty) \). - To find the range, we can calculate the y-coordinate of the vertex by plugging \( x = -\frac{5}{2} \) into the function. 3. \( K(x) = 2x^2 + 5x - 6 \) - This is another quadratic function with a positive coefficient for \( x^2 \), so it opens upwards. - The domain is all real numbers, \( (-\infty, \infty) \). - The range can be found by analyzing the vertex, which is at \( x = -\frac{b}{2a} \), where \( a = 2 \) and \( b = 5 \). 4. \( k(x) = \frac{1}{2}x^2 - \frac{2}{x}x \) - This function can be simplified to \( k(x) = \frac{1}{2}x^2 - 2 \) because \( \frac{2}{x}x = 2 \). - The domain is all real numbers except for \( x = 0 \) because division by zero is undefined. - The range can be found by analyzing the vertex of the parabola. Let's calculate the y-coordinates of the vertices for \( j(x) \) and \( K(x) \): For \( j(x) \): \[ y = j\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) - 6 \] \[ y = \frac{25}{4} - \frac{25}{2} - 6 \] \[ y = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} \] \[ y = \frac{25 - 50 - 24}{4} \] \[ y = \frac{-49}{4} \] For \( K(x) \): \[ y = K\left(-\frac{5}{4}\right) = 2\left(-\frac{5}{4}\right)^2 + 5\left(-\frac{5}{4}\right) - 6 \] \[ y = 2\left(\frac{25}{16}\right) - \frac{25}{4} - 6 \] \[ y = \frac{50}{16} - \frac{100}{16} - \frac{96}{16} \] \[ y = \frac{50 - 100 - 96}{16} \] \[ y = \frac{-146}{16} \] \[ y = -\frac{73}{8} \] Now, let's find the range for \( k(x) \): For \( k(x) = \frac{1}{2}x^2 - 2 \), the vertex is at \( x = 0 \) because the coefficient of \( x^2 \) is positive and the function is a parabola. The y-coordinate of the vertex is \( k(0) = \frac{1}{2}(0)^2 - 2 = -2 \). The range of \( k(x) \) is all real numbers less than or equal to -2, \( (-\infty, -2] \). In summary: - \( h(x) \) has a range of \( (-\infty, 0] \). - \( j(x) \) has a range of \( (-