Determine the coordinates of the vertex of the function f = 2x^2-5x+3.
f(x) = 2x^2 - 5x + 3
F(x) is a parabola where:
a= 2 b= -5 c=3
The coordinates of th vertex is:
(x, y) such that:
x= -b/2a = 5/4
y= (4ac-b^2)/4a = (4*2*3 - 25)/4 = -1/8
Ot to find y, we could substitute with x value:
f(5/4) = -1/8
Then the coordinates is:
To determine the coordinates of the vertex of the parable, we'll use the formula:
V (-b/2a ; -delta/4a)
We'll identify the coordinates a,b,c, of the expression of the function:
a = 2 , b = -5 , c = 3
Now, we'll calculate xV:
xV = 5/4
yV = -delta/4a
yV = (4ac-b^2)/4a
yV = (24-25)/8
yV = -1/8
The coordinates of the vertex are V(5/4 , -1/8).
f(x) = y = 2x^2-5x+3. to find the vertex.
y = 2x^2-5x+3
y/2 = x^2 -5x/2+3/2
y/2 = (x-5/4)^2 - (5/4)^2+3
y/2 = (x-5/4)^2 + (-25+48)/1= (x-5/4)^2 + (23/16)
y/2 -23/16) = (x-5/4)^2
(1/2) (y-23/8) = (x-5/4)^2. If we compare this parabola with the standard parabola 4aY = X^2 with vertex X = 0 and Y = 0, we get
x-5/4 = 0 and y-23/8 = 0 gives the coordinates of the given parabola.
So x= 5/4 and y = 23/8 are the coordinates of the vertex.