# How do you find the axis of symmetry, and the maximum or minimum value of the function v(t) = t^2 + 11t - 4?

Feb 3, 2018

$x = - \frac{11}{2} , \text{minimum at } - \frac{137}{4}$

#### Explanation:

$\text{for a quadratic in standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex which is also the axis}$
$\text{of symmetry is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$v \left(t\right) = {t}^{2} + 11 t - 4 \text{ is in standard form}$

$\text{with "a=1,b=11" and } c = - 4$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{11}{2}$

$\Rightarrow \text{equation of axis of symmetry is } x = - \frac{11}{2}$

$\text{substitute "x=-11/2" into equation for y}$

${y}_{\textcolor{red}{\text{vertex}}} = {\left(- \frac{11}{2}\right)}^{2} + 11 \left(- \frac{11}{2}\right) - 4 = - \frac{137}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{11}{2} , - \frac{137}{4}\right)$

$\text{the max/min value occurs at the vertex}$

$\text{for max/min consider the value of the coefficient a}$

• " if "a>0" then minimum value"

• " if "a<0" then maximum value"

$\text{here "a>0" hence minimum value}$

$\text{the minimum value } = - \frac{137}{4}$