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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

Respuesta :

merlynthewhizz

Function 1 [tex]f(x)=- x^{2} +4x+8[/tex]


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

Second step: Find the value of y-coordinate by substituting [tex]x=2[/tex] into [tex]f(x)[/tex] which give [tex]y=- (2)^{2} +4(2)+8=12[/tex]

Third step: Find the distance of the foci from the y-coordinate
[tex]y=- x^{2} +4x+8[/tex] - Multiply all term by -1 to get a positive [tex] x^{2} [/tex]
[tex]-y= x^{2} -4x-8[/tex] - then manipulate the constant of y to get a multiply of 4
[tex]4(- \frac{1}{4})y= x^{2} -4x-8 [/tex]
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is [tex]12- \frac{1}{4}=11.75 [/tex]

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

Function 2: [tex]f(x)= 2x^{2}+16x+18[/tex]

The function has a positive [tex] x^{2} [/tex] so it has a minimum

First step - [tex]x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4 [/tex]
Second step - [tex]y=2(-4)^{2}+16(-4)+18=-14 [/tex]
Third step - Manipulating [tex]f(x)[/tex] to leave [tex] x^{2} [/tex] with constant of 1
[tex]y=2 x^{2} +16x+18[/tex] - Divide all terms by 2
[tex] \frac{1}{2}y= x^{2} +8x+9 [/tex] - Manipulate the constant of y to get a multiply of 4
[tex]4( \frac{1}{8}y= x^{2} +8x+9 [/tex]

So the distance of focus from y-coordinate is [tex] \frac{1}{8} [/tex] to the north of [tex]y=-14[/tex]
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: [tex]f(x)=-2 x^{2} +5x+14[/tex]

First step: the function's maximum value happens when [tex]x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25 [/tex]
Second step: [tex]y=-2(1.25)^{2}+5(1.25)+14=17.125 [/tex]
Third step: Manipulating [tex]f(x)[/tex]
[tex]y=-2 x^{2} +5x+14[/tex] - Divide all terms by -2
[tex]-2y= x^{2} -2.5x-7[/tex] - Manipulate coefficient of y to get a multiply of 4
[tex]4(- \frac{1}{8})y= x^{2} -2.5x-7 [/tex]
So the distance of the foci from the y-coordinate is -[tex] \frac{1}{8} [/tex] south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

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

elpink25

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.

Ver imagen elpink25