Function 1

First step: Finding when

is minimum/maximum
The function has a negative value

hence the

has a maximum value which happens when

. The foci of this parabola lies on

.
Second step: Find the value of y-coordinate by substituting

into

which give

Third step: Find the distance of the foci from the y-coordinate

- Multiply all term by -1 to get a positive


- then manipulate the constant of y to get a multiply of 4

So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is

Hence the coordinate of the foci is (2, 11.75)
Function 2:

The function has a positive

so it has a minimum
First step -

Second step -

Third step - Manipulating
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to leave

with constant of 1

- Divide all terms by 2

- Manipulate the constant of y to get a multiply of 4
So the distance of focus from y-coordinate is

to the north of

Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)
Function 3:

First step: the function's maximum value happens when

Second step:

Third step: Manipulating


- Divide all terms by -2

- Manipulate coefficient of y to get a multiply of 4

So the distance of the foci from the y-coordinate is -

south to y-coordinate
Hence the coordinate of foci is (1.25, 17)
Function 4: following the steps above, the maximum value is when

and

. The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)
Function 5: the minimum value of the function is when

and

. Manipulating coefficient of y, the distance of foci from y-coordinate is

to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)
Function 6: The maximum value happens when

and

. The distance of the foci from the y-coordinate is

to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)