# How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. v(t) = t^2 + 11t - 4?

May 19, 2015

$v \left(t\right) = {t}^{2} + 11 t - 4.$
Coordinates of vertex: (x, y)

$x = \left(- \frac{b}{2} a\right) = \left(- \frac{11}{2}\right)$

$y = v \left(- \frac{11}{2}\right) = \frac{121}{4} - \frac{121}{2} - 4 = - \frac{137}{4}$

The function in vertex form:

$v \left(t\right) = {\left(x + \frac{11}{2}\right)}^{2} - \frac{137}{4}$

Check by developing:

$v \left(t\right) = {x}^{2} + 11 t + \frac{121}{4} - \frac{137}{4} = {t}^{2} + 11 t - 4 .$ OK