# Calculate the vertex of the graph of f(x)=x^2-5x+6. Determine the quadrant where the vertex is .

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To calculate the vertex of the quadratic, we'll apply the formula:

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

The quadratic function is:

f(x) = ax^2 + bx + c

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

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

Now, we'll calculate the coordinate xV:

xV = 5/1

**xV = 5**

yV = -delta/4a

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

yV = (24-25)/4

**yV = -1/4**

To determine the quadrant, we'll have to verify where the value of the coordinate x of a point is positive and where the value of the coordinate y is negative.

Since the positive values for x coordinate and negative values for y coordinate are found in the fourth quadrant, we'll conclude that: **The coordinates of the vertex of the parable are V(5, -1/4), so, according to the rule, they are located in the fourth quadrant, where xV>0 and yV<0.**

f(x) = x^2-5x+6. Or

y = x^2-5x+6. To calculate the vertex of the graph and determine the quadrant of the vertex.

Solution:

y = x^2-5x+6.

Rewrite this as:

y = x^2-5x+(5/2)^2 -(5/2)^2+6

y = (x-5/2)^2 + (24-25)/4

y = (x-5/2)^2 - 1/4

y+1/4 = (x-5/2)^2.........(1)

So this a parabola like the standard parabola X^2 = 4aY with vertex (X, Y) = (0 , 0 ).

So by comparision eq (1) is a parabola with vertax (x , y)= (5/2 , -1/4), which is in the 4th quadrant.