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What are the coordinates of the vertex of the parabola? 

y=3/4x^2−6x+15

 May 22, 2019
edited by Guest  May 22, 2019
 #1
avatar+1606 
+1

Hint: Use formula \(\frac{-b}{2a}\).

 

This is used to find the vertex, it should be trivial

 May 22, 2019
 #2
avatar+103982 
+2

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

 

 

cool cool cool

 May 22, 2019

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