Determine the function f=x^2-ax+3 whose minim value is -1.

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