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

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To find the vertex of a parabola, we can look for points where the first derivative is equal to zero:

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

=> f'(x) = 4x - 7

Now we set f'(x) equal to zero and solve:

4x - 7 = 0

=> 4x = 7

=> x = 7/4

Now we know that the x-coordinate of the vertex is at 7/4, so lets plug that back into our function to find the y-coordinate:

f(7/4) = 2*(7/4)^2 - 7*(7/4) + 5

= -9/8

Therefore,

**The vertex of the function f(x) = 2x^2 -7x + 5 is located at the point (7/4, -9/8)**

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

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

The minimum value of parabola 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 represented by the pair: (7/4 ; -9/8).**