Question

Match the parabolas represented by the equations with their foci.

1. Y=-x^2+4x+8
2. Y=2x^2+16x+18
3. Y=-2x^2+5x+14
4. Y=-x^2+17x+7
5. Y=2x^2+11x+5
6.y=-2x^2+6x+5
-Pairs-
A. (-2.75, -10)
B. (2, 11.75)
C. (-4, -13.875)
D. (1.25, 17)

Please match numbers 1-6 with letters a-d I really need help thanks

Answer 1

Answer: 1) (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

Step-by-step explanation: I got this right on Edmentum.

Answer 2

Function 1
`f(x)=- x^(2) +4x+8`


First step: Finding when
`f(x)` is minimum/maximum
The function has a negative value
`x^(2)` hence the
`f(x)` has a maximum value which happens when
`x=- (b)/(2a)=- (4)/((2)(1))=2`. The foci of this parabola lies on
`x=2`.

Second step: Find the value of y-coordinate by substituting
`x=2` into
`f(x)` which give
`y=- (2)^(2) +4(2)+8=12`

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

`y=- x^(2) +4x+8` - Multiply all term by -1 to get a positive
`x^(2)`

`-y= x^(2) -4x-8` - then manipulate the constant of y to get a multiply of 4

`4(- (1)/(4))y= x^(2) -4x-8`
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is
`12- (1)/(4)=11.75`

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

Function 2:
`f(x)= 2x^(2)+16x+18`

The function has a positive
`x^(2)` so it has a minimum

First step -
`x=- (b)/(2a)=- (16)/((2)(2))=-4`
Second step -
`y=2(-4)^(2)+16(-4)+18=-14`
Third step - Manipulating
`f(x)` to leave
`x^(2)` with constant of 1

`y=2 x^(2) +16x+18` - Divide all terms by 2

`(1)/(2)y= x^(2) +8x+9` - Manipulate the constant of y to get a multiply of 4

`4( (1)/(8)y= x^(2) +8x+9`

So the distance of focus from y-coordinate is
`(1)/(8)` to the north of
`y=-14`
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3:
`f(x)=-2 x^(2) +5x+14`

First step: the function's maximum value happens when
`x=- (b)/(2a)=- (5)/((-2)(2))= (5)/(4)=1.25`
Second step:
`y=-2(1.25)^(2)+5(1.25)+14=17.125`
Third step: Manipulating
`f(x)`

`y=-2 x^(2) +5x+14` - Divide all terms by -2

`-2y= x^(2) -2.5x-7` - Manipulate coefficient of y to get a multiply of 4

`4(- (1)/(8))y= x^(2) -2.5x-7`
So the distance of the foci from the y-coordinate is -
`(1)/(8)` south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when
`x=8.5` and
`y=79.25`. 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
`x=-2.75` and
`y=-10.125`. Manipulating coefficient of y, the distance of foci from y-coordinate is
`(1)/(8)` 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
`x=1.5` and
`y=9.5`. The distance of the foci from the y-coordinate is
`(1)/(8)` to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)