1. **Problem:** Find the vertex form, y-intercept, axis of symmetry, vertex, domain, and range of the function $$y = \frac{1}{2}x^2 + x - 4$$. 2. **Formula and rules:** - The vertex form of a quadratic function is $$y = a(x-h)^2 + k$$ where $$(h,k)$$ is the vertex. - The axis of symmetry is the vertical line $$x = h$$. - The y-intercept is found by evaluating $$y$$ at $$x=0$$. - The domain of any quadratic function is all real numbers, $$(-\infty, \infty)$$. - The range depends on the vertex and the direction of the parabola (upward if $$a>0$$, downward if $$a<0$$). 3. **Find the vertex:** - Use $$h = -\frac{b}{2a}$$ where $$a=\frac{1}{2}$$ and $$b=1$$. - Calculate $$h = -\frac{1}{2 \times \frac{1}{2}} = -\frac{1}{1} = -1$$. - Find $$k = y(h) = \frac{1}{2}(-1)^2 + (-1) - 4 = \frac{1}{2} - 1 - 4 = -4.5$$. - So, vertex is $$(-1, -4.5)$$. 4. **Vertex form:** - Substitute $$h$$ and $$k$$ into vertex form: $$y = \frac{1}{2}(x + 1)^2 - 4.5$$. 5. **Y-intercept:** - Evaluate $$y$$ at $$x=0$$: $$y = \frac{1}{2}(0)^2 + 0 - 4 = -4$$. 6. **Axis of symmetry:** - The line $$x = -1$$. 7. **Domain:** - All real numbers, $$(-\infty, \infty)$$. 8. **Range:** - Since $$a=\frac{1}{2} > 0$$, parabola opens upward. - Minimum value is $$k = -4.5$$. - So, range is $$[-4.5, \infty)$$. **Final answer:** - Vertex form: $$y = \frac{1}{2}(x + 1)^2 - 4.5$$ - Y-intercept: $-4$ - Axis of symmetry: $$x = -1$$ - Vertex: $$(-1, -4.5)$$ - Domain: $$(-\infty, \infty)$$ - Range: $$[-4.5, \infty)$$