# Determine the vertex of the function 2x^2-7x+5

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### 3 Answers

f(x) = 2x^2 - 7x + 5

we know that:

**a= 2 b = -7 c = 5**

We also know that the vertix V is:

v ( vx , vy ) such that:

vx= -b/2a

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

Let us calculate:

vx= -b/2a = 7/2*2 = 7/4

vy = - (b^2 - 4ac)/4a = -(49 - 4*2*5)/4*2

= -9/8

==> vy = -9/8

Then the vertix v is:

**V ( 7/4, -9/8)**

Let the function be y(x) = 2x^2-7x+5.

To determine the vertex, we first bring this into the standard form of a parabola like (x-h)^2 = 4a(y-k), where (h,k) are the vertex of the parabola, a is focal distance from the origin along the y axis.

y = 2x^2-7x +5 .

We divide both sides by 2.

y/2 = x^2-(7/2)x+5/2.

We write right as (x^2-7/4 )^2 - (7/4)^2 +5/2.

y/2 = (x-7/4)^2 - 49/16 +40/16.

y/2 = (x-7/4)^2 -9/16.

y/2 +9/16 = (x-7/4)^2.

(y + 9/8)/2 = (x-7/4)^2.

(x-7/4)^2 = (y+9/8)/2.

(x-7/4)^2 = 4*(1/8)(y+9/8).

Comparing the above with the standard parabola (x-h)^2= 4a(y-k), we see that the coordinates of the vetex (h, k) = (7/4 , -9/8).

Since the expression of the function is a quadratic, we'll consider the vertex of the parable as the extreme point.

Since the coefficient of x^2 is positive, the vertex is a minimum point.

f(x) = 2x^2-7x+5

The minimum value is V(-b/2a , -delta/4a)

We'll identify the coefficients a,b,c:

a = 2

b = -7

c = 5

delta = b^2 - 4ac

delta = 49 - 40

delta = 9

The vertex V has the following coordinates:

xV = -b/2a

xV = -(-7) / 2*2

**xV = 7/4**

yV = -9/4*2

**yV = -9/8**

**The coordinates of the vertex of the function are: (7/4 ; -9/8).**