# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of v(t) = t^2 + 11t - 4?

Sep 29, 2015

#### Explanation:

$f \left(x\right) = a {x}^{2} + b x + c \implies$the general/standard form of the quadratic function, it is represented by the graph of the parabola where the coordinate of the vertex is:
$\left[- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right]$ , so in this case we have:
$v \left(t\right) = {t}^{2} + 11 t - 4$ , then:
$t = - \frac{11}{2}$
$v \left(- \frac{11}{2}\right) = {\left(- \frac{11}{2}\right)}^{2} + 11 \left(- \frac{11}{2}\right) - 4$
$= \frac{121}{4} - \frac{121}{2} - 4 = - \frac{137}{4}$
$\therefore$ the vertex is at $\left(- \frac{11}{2} , - \frac{137}{4}\right)$

The axis of symmetry is the same as the x-coordinate of the vertex, so in this case:
$t = - \frac{11}{2} \implies$ axis of symmetry

To find the y-intercept set x to zero and solve for y, in this case:
$v = 0 + 0 - 4 = - 4 \implies$the y-intercept is at $\left(0 , - 4\right)$

To find the x-intercepts set y to zero and solve for x:
${t}^{2} + 11 t - 4 = 0 \implies$ solution by completing the square:
${t}^{2} + 11 t + {\left(\frac{11}{2}\right)}^{2} = 4 + \frac{121}{4}$
${\left(t + \frac{11}{2}\right)}^{2} = \frac{137}{4}$
$t + \frac{11}{2} = \pm \frac{\sqrt{137}}{2}$
$t = - \frac{11}{2} \pm \frac{\sqrt{137}}{2} \implies$ hence the x-intercepts are:
$\left(- \frac{11}{2} \pm \frac{\sqrt{137}}{2} , 0\right)$

The domain of parabola is all the real numbers, but the range is limited by y-coordinates of the vertex, in this case since the parabola opens up the vertex is the minimum so the range is:
$v \ge - \frac{137}{4}$