# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=3(x-3)^2+4?

Jun 4, 2015

graph{3x^2 - 18x + 31 [-10, 10, -5, 5]}
This function y is written the vertex form

x of vertex: x = 3

y of vertex y = 4

Axis of symmetry: x = 3

x- intercepts -> Find the standard form

$y = 3 \left({x}^{2} - 6 x + 9\right) + 4 = 0$
$y = 3 {x}^{2} - 18 x + 31$
$D = {d}^{2} = {b}^{3} - 4 a c = 324 - 372 = - 48 < 0.$
There are no real roots (no x- intercepts). The graph is completely above the x axis