Question: 2.) f(x)=x²+x+6
3.) f(x)=-x²+2x+1
4.) f(x)=x²-4x+3
solve/find the ff:
•vertex form
•vertex
•axis of symmetry
•y-intercept
•x-intercept
1. **State the problem:** We are given the quadratic function $$f(x)=x^2+x+6$$ and need to find its vertex form, vertex, axis of symmetry, y-intercept, and x-intercepts.
2. **Recall the vertex form of a quadratic:**
$$f(x)=a(x-h)^2+k$$ where $$(h,k)$$ is the vertex.
3. **Find the vertex form by completing the square:**
$$f(x)=x^2+x+6$$
Group the first two terms:
$$= (x^2 + x) + 6$$
Take half the coefficient of $$x$$, which is $$\frac{1}{2}$$, square it: $$\left(\frac{1}{2}\right)^2=\frac{1}{4}$$.
Add and subtract $$\frac{1}{4}$$ inside the parentheses:
$$= \left(x^2 + x + \frac{1}{4} - \frac{1}{4}\right) + 6$$
Rewrite:
$$= \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + 6$$
Simplify constants:
$$= \left(x + \frac{1}{2}\right)^2 + \frac{23}{4}$$
4. **Vertex:**
From vertex form, vertex is $$\left(-\frac{1}{2}, \frac{23}{4}\right)$$.
5. **Axis of symmetry:**
The axis is the vertical line through the vertex:
$$x = -\frac{1}{2}$$
6. **Y-intercept:**
Set $$x=0$$:
$$f(0) = 0^2 + 0 + 6 = 6$$
So y-intercept is $$6$$.
7. **X-intercepts:**
Set $$f(x)=0$$:
$$x^2 + x + 6 = 0$$
Calculate discriminant:
$$\Delta = b^2 - 4ac = 1^2 - 4(1)(6) = 1 - 24 = -23$$
Since $$\Delta < 0$$, 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)