# Calculate in what quadrant is the vertex of the graph of f=x^2-5x+6

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To determine the quadrant, we'll have to calculate the coordinates of the vertex of the parable.

We'll apply the formula:

V (-b/2a ; -delta/4a)

We'll identify the coordinates a,b,c, of the expression of the function:

a =1 , b = -5, c = 6

Now, we'll calculate xV:

xV = 5/1

xV = 5

yV = -delta/4a

yV = (4ac-b^2)/4a

yV = (24-25)/4

yV = -1/4

**The coordinates of the vertex of the parable are V(5, -1/4) and they are located in the fourth quadrant, where xV>0 and yV<0. **

The vertex of a parabola y = ax^2+by+c is

given by: (xV,yV ) = (-b/2a , -(b^2-4ac)/4a)....(1)

So the given parabola is y = x^2-5x+6

a = 1, b = -5 and c = 6

Therefore substituting in (1)

(xV , yV) ={ - -5/2*1 , ((-5)^2 -4*1+6))/4*1 }

= (5, 1/4) which is in 1st quadrant.