Quadratic Properties A7F250

1. **Problem:** Find the vertex form, vertex, axis of symmetry, y-intercept, and x-intercept of the quadratic function $$f(x)=x^2+x+6$$. 2. **Formula and rules:** - Vertex form of a quadratic: $$f(x)=a(x-h)^2+k$$ where $$(h,k)$$ is the vertex. - Vertex coordinates: $$h=-\frac{b}{2a}$$ and $$k=f(h)$$. - Axis of symmetry: $$x=h$$. - Y-intercept: value of $$f(0)$$. - X-intercepts: solve $$f(x)=0$$. 3. **Calculate vertex:** - Here, $$a=1$$, $$b=1$$, $$c=6$$. - $$h=-\frac{1}{2(1)}=-\frac{1}{2}$$. - $$k=f\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)+6=\frac{1}{4}-\frac{1}{2}+6=\frac{1}{4}-\frac{2}{4}+\frac{24}{4}=\frac{23}{4}$$. 4. **Vertex form:** - $$f(x)=1\left(x-\left(-\frac{1}{2}\right)\right)^2+\frac{23}{4}=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}$$. 5. **Axis of symmetry:** - $$x=-\frac{1}{2}$$. 6. **Y-intercept:** - $$f(0)=0^2+0+6=6$$. 7. **X-intercepts:** - Solve $$x^2+x+6=0$$. - Discriminant $$\Delta=b^2-4ac=1^2-4(1)(6)=1-24=-23<0$$. - No real x-intercepts. **Final answers:** - Vertex form: $$f(x)=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}$$ - Vertex: $$\left(-\frac{1}{2},\frac{23}{4}\right)$$ - Axis of symmetry: $$x=-\frac{1}{2}$$ - Y-intercept: $$6$$ - X-intercepts: None (no real roots)