# How do you graph a quadratic function without plugging in points y= 2(x-3)^2+4?

Apr 14, 2015

Basically you need:
1] The $y$ axis intercept; obtained setting $x = 0$
2] The vertex; obtained as:
${x}_{v} = - \frac{b}{2 a}$
${y}_{v} = - \frac{\Delta}{4 a} = - \frac{{b}^{2} - 4 a c}{4 a}$
When your function is in the form $y = a {x}^{2} + b x + c$
3] Possible intercepts on the $x$ axis; obtained setting $y = 0$.