Axis of Symmetry Vertex Y-intercept

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1. **Axis of Symmetry**

For a quadratic function in the form  
   $$
   f(x)=ax^2+bx+c
   $$
   the axis of symmetry is given by the formula  
   $$
   x=-\frac{b}{2a}.
   $$

2. **Vertex**

The vertex of the parabola is the point where the function attains its minimum (if $$a>0$$) or maximum (if $$a<0$$). Its $$x$$-coordinate is the same as that of the axis of symmetry, i.e.,  
   $$
   x=-\frac{b}{2a}.
   $$  
   To find the $$y$$-coordinate of the vertex, substitute this value into the function:  
   $$
   y=f\left(-\frac{b}{2a}\right)=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c.
   $$  
   Thus, the vertex is located at  
   $$
   \left(-\frac{b}{2a},\, a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c\right).
   $$

3. **Y-intercept**

The y-intercept is the point where the graph of the function crosses the $$y$$-axis, which happens when $$x=0$$. Substituting $$x=0$$ into the function gives:  
   $$
   y=f(0)=a(0)^2+b(0)+c=c.
   $$  
   Therefore, the y-intercept is  
   $$
   (0,\,c).
   $$
- **Axis of Symmetry**: $$ x = -\frac{b}{2a} $$ - **Vertex**: $$ \left( -\frac{b}{2a}, \, a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c \right) $$ - **Y-intercept**: $$ (0, \, c) $$

Axis of Symmetry Vertex Y-intercept

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