How do you find the vertex, axis of symmetry and intercepts of f(x)=3x^2+x-3?

Feb 15, 2016

f(x) = 3x^2 + x - 3

Explanation:

x-coordinate of vertex, or axis of symmetry:
$x = - \frac{b}{2 a} = - \frac{1}{6}$
y-coordinate of vertex:
$f \left(- \frac{1}{6}\right) = 3 \left(\frac{1}{36}\right) - \frac{1}{6} - 3 = \frac{1}{12} - \frac{1}{6} - 3 = - \frac{17}{6}$
To find y-intercept, make x = 0 --> y = -3
To find x-intercepts, make y = 0 and solve the quadratic equation
$y = 3 {x}^{2} + x - 3 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 1 + 36 = 37$ -->$d = \pm \sqrt{37}$
There are 2 real roots, or two x-intercepts:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{6} \pm \frac{\sqrt{37}}{6}$