Question

The vertex of the quadratic expression $y=3x^2+2x+5$ is given as:

• A. $(\frac{14}{3},-\frac{1}{3})$
• B. $(-\frac{14}{3},\frac{1}{3})$
• C. $(-\frac{1}{3},\frac{14}{3})$
• D. $(\frac{1}{3},-\frac{14}{3})$

Answer 1

The correct option is C $(-\frac{1}{3},\frac{14}{3})$

Given quadratic expression: $y=3x^2+2x+5$
On comparing with the standard form of quadratic expression $y=a{x}^2+bx+c,$ we get:
$a=3,b=2,c=5\&D=b^2-4ac=(2{)}^2-4\cdot 3\cdot 5=-56$
$a>0\Rightarrow$ The parabola is upward opening.
The coordinates of the vertex will be given as:
$(\frac{-b}{2a},\frac{-D}{4a})$
Substituting the values of $a,b\&D$ in the equation, we get the coordinates of the vertex as:
$V\equiv (\frac{-b}{2a},\frac{-D}{4a})\equiv (\frac{-2}{6},-\frac{(-56)}{12})\equiv (\frac{-1}{3},\frac{14}{3})$

$\Rightarrow V\equiv (\frac{-1}{3},\frac{14}{3})$