Jawaban:

Setelah menyimak penjelasan singkat diatas, mari kita selesaikan soal tersebut.

✈ Diketahui :

y = {x}^{2} + 4x - 12y=x

2

+4x−12

✈ Ditanya :

Tentukan titik potong sumbu x, titik potong sumbu y, sumbu simetri, titik puncak dan titik bantu...?

✈ Jawab :

y = {x}^{2} + 4x - 12y=x

2

+4x−12

☛ Titik potong sumbu x ( y = 0 )

{x}^{2} + 4x - 12 = 0x

2

+4x−12=0

( x + 6)( x - 2) = 0(x+6)(x−2)=0

x = -6 \: \: atau \: \: x = 2x=−6ataux=2

Titik \: potong : {\red{\boxed{\bf{ \red{(-6,0) \: dan \: (2,0)}}}}}Titikpotong:

(−6,0)dan(2,0)

☛ Titik potong dengan sumbu y ( x = 0 )

y = {x}^{2} + 4x - 12y=x

2

+4x−12

y = {0}^{2} + 4.0 - 12y=0

2

+4.0−12

y = -12y=−12

Titik \: potong : {\red{\boxed{\bf{ \red{(0,-12)}}}}}Titikpotong:

(0,−12)

☛ Sumbu simetri

xs = \frac{ - b}{2a}xs=

2a

−b

= \frac{ - 4}{2 \times 1}=

2×1

−4

= {\red{\boxed{\bf{ \red{-2}}}}}=

−2

☛ Titik optimum

ys = {x}^{2} + 4x - 12ys=x

2

+4x−12

ys = ({ - 2})^{2} + 4(-2) - 12ys=(−2)

2

+4(−2)−12

= 4 - 8 - 12=4−8−12

= {\red{\boxed{\bf{ \red{-16}}}}}=

−16

☛ Titik puncak

={\red{\boxed{\bf(-2,-16)}}}=

(−2,−16)

☛ Titik bantu

1. \: { \bf{x = -7}}1.x=−7

y = ( - 7)^{2} + 4(-7) - 12y=(−7)

2

+4(−7)−12

= 49 - 28 - 12=49−28−12

= 9=9

= = > {\red{\boxed{\bf{ \red{(-7,9)}}}}}==>

(−7,9)

2. \: { \bf{x = 3}}2.x=3

y = {3}^{2} + 4(3) - 12y=3

2

+4(3)−12

= 9 + 12 - 12=9+12−12

= 9=9

= = > {\red{\boxed{\bf{ \red{(3,9)}}}}}==>

(3,9)

3. \: { \bf{x = -4}}3.x=−4

y = ( - 4)^{2} + 4(-4) - 12y=(−4)

2

+4(−4)−12

= 16 - 16 - 12=16−16−12

= -12=−12

= = > {\red{\boxed{\bf{ \red{(-4,-12)}}}}}==>

(−4,−12)