Question

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

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

Answer 1

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

Given quadratic expression: $y=3x^2+2x+5$
On comparing with standard form of quadratic expression $y=a{x}^2+bx+c$
we get, $a=3,b=2,c=5\&D=b^2-4ac=(2{)}^2-\mathrm{4.3.5}=-56$
Here, $a>0\Rightarrow$ The parabola is upward opening.
The coordinates of the vertex will be given as:
$(\frac{-b}{2a},\frac{-D}{4a})$
Substituting the values of $a,b\&c$ in the equation, we get:
$(\frac{-b}{2a},\frac{-D}{4a})\equiv (\frac{-2}{6},-\frac{(-56)}{12})\equiv (\frac{-1}{3},\frac{14}{3})$
Hence, the coordinates of vertex are given by:
$(\frac{-1}{3},\frac{14}{3})$