What are the coordinates of the vertex of the parabola?
y=3/4x^2−6x+15
Hint: Use formula \(\frac{-b}{2a}\).
This is used to find the vertex, it should be trivial
y = (3/4)x^2 - 6x + 15
We have the form y = Ax^2 + Bx + C
The x coordinate of the vertex is given by -B/[2A] = - -6 / [ 2(3/4)] = 6 / [ 6/4] = 6 *4/6 = 4
So....the y coordinate of the vertex is g(3/4)(4)^2 - 6(4) + 15 = (3/4)16 - 24 + 15 = 12 - 24 + 15 = 3
So....the vertex is ( 4, 3)
Here's a graph : https://www.desmos.com/calculator/vkss4gjgpf