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

$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=$$-\frac{b}{2a}$$
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}$