f(x) = x^2 - ax + 3

The function has a minimuum value = -1

==> f(x1) = -1 where x1 is the derivative's root

let us differentiate f(x):

f'(x) = 2x - a = 0

==> 2x = a

==> x1= a/2

Then a/2 is is a critical value.

the function f has a minimum value when x= a/2.

But f(a/2) = -1

==> (a/2)^2 - a*a/2 + 3 = -1

==> a^2/4 - a^2/2 + 4 = 0

Multiply by 4;

==> a^2 - 2a^2 + 16 = 0

==> -a^2 + 16 = 0

==> a^2 = 16

**Then there are two solutions:**

**==> a1= 4**

**==> a2= -4**

The extreme value of a quadratic function can be calculated using 2 methods.

We'll apply the first method of finding the minimum value. We'll differentiate the function.

f'(x) = (x^2 - a*x + 3)'

f'(x) = (x^2)' - (a*x)' + (3)'

f'(x) = 2x - a

When the first derivative is cancelling, then the function has an extreme value, in this case, a minimum.

2x - a = 0

We'll add a both sides:

2x = a

We'll divide by 2:

x = a/2

But, from enunciation, f(a/2) = -1.

We'll determine f(a/2):

f(a/2) = a^2/4 - a^2/2 + 3

-1 = (a^2 - 2a^2 + 12)/4

-4 = -a^2 + 12

We'll move -a^2 to the left side:

a^2 - 4 = 12

We'll add 4 both sides:

a^2 = 16

a1 = -sqrt 16

**a1 = -4**

**a2 = +4**

The second method of finding the extreme value of the quadratic is to apply the formula of the vertex of the graph:

V (-b/2a , -delta/4a)

We'll identify the coefficients of the function:

a = 1

b = -a

c = 3

xV = -(-a)/2

delta = a^2 - 12

yV = -(a^2 - 12)/4

But yV= -1

-(a^2 - 12)/4 = -1

a^2 - 12 = 4

a^2 = 12+4

a^2 = 16

a1 = 4

a2 = -4

f(x) = x^2-ax+3.

The minimum value is -1.

To determine the function or determine a in x^2-ax+3.

Solution:

f(x) = x^2-ax+3 = (x-a/2)^2 - (a/2)^2. has the minimum when x = a/2..

Therefore min f(x) = f(a/2) = (a/2-/2)^2 -(a/2)^2+3 which should be equal to -1.

Or

min f(x) = 0 -(a/2)^2+3 = -1

-a^2/4 +3 = -1

-a^2 +12 = -4.

a^2 -12 = 4

a^2 = 4+12 = 16.

a = -4 or 4.

Therefore the function is f(x) = x^2-4x+3 which is minimum when x = 2.

Or the function is f(x) = x^2+4x+3 which is minimum when x = -2.