Calculate the vertex of the graph of f(x)=x^2-5x+6. Determine the quadrant where the vertex is .
f(x) = x^2 - 5x + 6
==> a= 1 b = -5 c= 6
The coordinates of the vertex is:
x = -b/a = 5/1 = 5
y= -(b^2 - 4ac)/4a = - (25 - 4*1*6) / 4 = -1/4
Then the coordinates of the vertix is:
V = (5, -1/4)
x >0 and y < 0, then the point V is in the 4th quadrant.
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.
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.