# What is the vertex of  y=-3x^2 + 2x - 5?

The Vertex is at ($\frac{1}{3} , - 4 \frac{2}{3}$)
This is the equation of the Parabola opens down as co-efficient of ${x}^{2}$ is negative. Comparing with the General equation ($a {x}^{2} + b x + c$)
we get a = (-3) ; b= 2 ; c =(-5) Now we know x-co-ordinate of the vertex is equal to -b/2a. so ${x}_{1} = - \frac{2}{2 \cdot \left(- 3\right)}$ or ${x}_{1} = \frac{1}{3}$
Now putting the value of x =1/3 in the equation we get ${y}_{1} = - 3. {\left(\frac{1}{3}\right)}^{2} + 2 \cdot \left(\frac{1}{3}\right) - 5$ or ${y}_{1} = - \frac{14}{3}$ or ${y}_{1} = - \left(4 \frac{2}{3}\right)$ So The Vertex is at$\left(\frac{1}{3} , - 4 \frac{2}{3}\right)$