How do you identify the important parts of #y= 6x^2 +2x+4# to graph it?

1 Answer
Oct 3, 2015

Graph #y = 6x^2 + 2x + 4#

Explanation:

The important parts to graph y are:
a. x-coordinate of vertex and axis of symmetry:
#x = (-b/2a) = -2/12 = -1/6#
y-coordinate of vertex:
#y = f(-1/6) = 1/6 - 2/6 + 4 = 23/6.#
b. y-intercept. Make x = 0 --> y = 4
x-intercepts. Make y = 0 and solve #6x^2 + 2x + 4 = 0.#
Since D = 4 - 96 < 0, there are no x-intercepts (no real roots). The parabola is completely above the x-axis and it opens upward (a > 0).
graph{6x^2 + 2x + 4 [-40, 40, -20, 20]}