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

1 Answer
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(x^2 - 6x + 9) + 4 = 0
y = 3x^2 - 18x + 31
D = d^2 = b^3 - 4ac = 324 - 372 = -48 < 0.
There are no real roots (no x- intercepts). The graph is completely above the x axis