1. **Problem:** Find the domain of the function $$y=\frac{x+1}{\sqrt{x-1}}$$. - The expression inside the square root must be positive because the denominator cannot be zero or negative inside a square root. - So, we require $$x-1>0$$ which gives $$x>1$$. - Therefore, the domain of $$y$$ is all $$x$$ such that $$x>1$$. 2. **Problem:** Find the equation of the line passing through the points $$(-1,-2)$$ and $$(3,-1)$$. - First, find the slope $$m$$: $$m=\frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-2)}{3 - (-1)} = \frac{-1 + 2}{3 + 1} = \frac{1}{4}$$. - Use point-slope form $$y - y_1 = m(x - x_1)$$ with point $$(-1,-2)$$: $$y - (-2) = \frac{1}{4}(x - (-1))$$ which simplifies to $$y + 2 = \frac{1}{4}(x + 1)$$ or $$y = \frac{1}{4}x + \frac{1}{4} - 2 = \frac{1}{4}x - \frac{7}{4}$$. 3. **Problem:** Graph the function $$y = -2x^2 + 5$$. - This is a downward opening parabola because the coefficient of $$x^2$$ is negative. - The vertex is at $$x=0$$ since no $$x$$ term is present. - Calculate vertex coordinate: $$y = -2(0)^2 + 5 = 5$$ - The vertex is at $$(0, 5)$$. - The parabola opens downward because the coefficient of $$x^2$$ is $$-2$$. 4. **Problem:** Graph the function $$y=3x^2 + 10x - 7$$. - This is a parabola opening upwards since the coefficient of $$x^2$$ is positive. - Find vertex using formula $$x = -\frac{b}{2a}$$ where $$a=3$$ and $$b=10$$: $$x = -\frac{10}{2\times3} = -\frac{10}{6} = -\frac{5}{3}$$. - Plug into the function to find $$y$$-coordinate of the vertex: $$y = 3\left(-\frac{5}{3}\right)^2 + 10\left(-\frac{5}{3}\right) - 7 = 3\times\frac{25}{9} - \frac{50}{3} - 7 = \frac{75}{9} - \frac{150}{9} - \frac{63}{9} = \frac{75 - 150 -63}{9} = \frac{-138}{9} = -\frac{46}{3}$$. - So the vertex is at $$\left(-\frac{5}{3}, -\frac{46}{3}\right)$$. - The parabola opens upward since $$a=3 > 0$$.