Find vertex and axis of symmetry from the quadratic equation $y=ax^2+bx+c$
Also, derive the quadratic formula.

Question

anonymous · Answer

$$y=ax^2 + bx + c $$$$y = a(x^2+\frac{b}{a}x)+c$$$$y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2)+c$$$$y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$$$$y=a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}$$
So, vertex is at $( -\frac{b}{2a} , -\frac{b^2-4ac}{4a})$, axis of symmetry is x=$ -\frac{b}{2a}$

anonymous · Answer

Deriving quadratic formula
To find x-intercepts, we put y=0
$$ax^2 + bx + c=0$$$$x^2 + \frac{b}{a}x + \frac{c}{a}=0$$$$x^2 + \frac{b}{a}x + (\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}=0$$$$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$$$$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$$$$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$$$$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$$$$x=\pm \frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}$$$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$