# Determine the coordinates of the vertex of the function f = 2x^2-5x+3.

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### 3 Answers

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:

(5/4, -1/8)

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.

Solution:

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.