Examples
f(x)=x2-6x+5
Step 1
The minimum of a quadratic function occurs at x=-b2a. If a is positive, the minimum value of the function is f(-b2a).
fminx=ax2+bx+c occurs at x=-b2a
Step 2
Step 2.1
Substitute in the values of a and b.
x=--62(1)
Step 2.2
Remove parentheses.
x=--62(1)
Step 2.3
Simplify --62(1).
Step 2.3.1
Cancel the common factor of -6 and 2.
Step 2.3.1.1
Factor 2 out of -6.
x=-2⋅-32⋅1
Step 2.3.1.2
Cancel the common factors.
Step 2.3.1.2.1
Factor 2 out of 2⋅1.
x=-2⋅-32(1)
Step 2.3.1.2.2
Cancel the common factor.
x=-2⋅-32⋅1
Step 2.3.1.2.3
Rewrite the expression.
x=--31
Step 2.3.1.2.4
Divide -3 by 1.
x=--3
x=--3
x=--3
Step 2.3.2
Multiply -1 by -3.
x=3
x=3
x=3
Step 3
Step 3.1
Replace the variable x with 3 in the expression.
f(3)=(3)2-6⋅3+5
Step 3.2
Simplify the result.
Step 3.2.1
Simplify each term.
Step 3.2.1.1
Raise 3 to the power of 2.
f(3)=9-6⋅3+5
Step 3.2.1.2
Multiply -6 by 3.
f(3)=9-18+5
f(3)=9-18+5
Step 3.2.2
Simplify by adding and subtracting.
Step 3.2.2.1
Subtract 18 from 9.
f(3)=-9+5
Step 3.2.2.2
Add -9 and 5.
f(3)=-4
f(3)=-4
Step 3.2.3
The final answer is -4.
-4
-4
-4
Step 4
Use the x and y values to find where the minimum occurs.
(3,-4)
Step 5