# How do you find the vertex of g(x)= 5/7x^2+ 13/21x + 1/3?

May 4, 2016

color(blue)("Vertex "->(x,y)->(-13/30,+251/1260)" Exactly"

color(blue)("Vertex "->(x,y)->(-0.43bar3, 0.199)" Approximately"

#### Explanation:

color(blue)("Determine "x_("vertex"))

Write as $\text{ } y = \frac{5}{7} \left[{x}^{2} + \left(\frac{13}{21} \div \frac{5}{7}\right) x\right] + \frac{1}{3}$

$y = \frac{5}{7} \left({x}^{2} + \frac{13}{15} x\right) + \frac{1}{3}$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{13}{15} = - \frac{13}{30}}$

'.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.
color(blue)("Determine "y_("vertex"))

Substitute for $x$ in the original equation

${y}_{\text{vertex}} = \frac{5}{7} {\left(- \frac{13}{30}\right)}^{2} + \frac{13}{21} \left(- \frac{13}{30}\right) + \frac{1}{3}$

${y}_{\text{vertex}} = \left(\frac{5}{7} \times \frac{160}{900}\right) - \left(\frac{13}{21} \times \frac{13}{30}\right) + \frac{1}{3}$

${y}_{\text{vertex}} = \frac{169}{1260} - \frac{169}{630} + \frac{1}{3}$

$\textcolor{b l u e}{{y}_{\text{vertex}} = \frac{251}{1260} \approx 0.1992}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

color(blue)("Vertex "->(x,y)->(-13/30,+251/1260)" Exactly"
$\textcolor{red}{\text{Thank goodness for calculators!}}$

color(blue)("Vertex "->(x,y)->(-0.43bar3, 0.199)" Approximately"