Question

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

Answer 1

The Vertex is at (`1/3, -4 2/3`)

Explanation:

This is the equation of the Parabola opens down as co-efficient of `x^2` is negative. Comparing with the General equation (`ax^2+bx+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=-2/(2*(-3))` or `x_1 = 1/3`
Now putting the value of x =1/3 in the equation we get `y_1=-3.(1/3)^2 +2*(1/3) -5` or `y_1 = -14/3` or `y_1=-(4 2/3)` So The Vertex is at`(1/3, -4 2/3)`