f(x) = x² + 5x + 4
[tex] ~ [/tex]
x² + 5x + 4 = 0
(x + 4)(x + 1) = 0
x = {-4, -1}
Koordinat (-4,0) dan (-1,0)
maka x = 0
y = x² + 5x + 4
y = 0² + 5(0) + 4
y = 4
Koordinat (0,4)
[tex] \tt = \frac{ - b}{2a} [/tex]
[tex] \tt = \frac{ - 5}{2(1)} [/tex]
[tex] \tt = \frac{ - 5}{2} [/tex]
= -2,5
[tex] \tt = \frac{ - b}{2a} , \frac{ - d}{4a} [/tex]
[tex] \tt = ( - \frac{5}{2} , - \frac{ - ( {5}^{2} - 4(1)(4)) }{4(1)} )[/tex]
[tex] \tt = ( - \frac{5}{2} , \frac{ - (25 - 16)}{4} [/tex]
[tex] \tt = ( - \frac{5}{2} , \frac{ - 9}{4} )[/tex]
[tex] \tt = ( - \frac{ 5}{2} , - \frac{9}{4} )[/tex]
= (-2.5 , -2.25)
Tarik garis pada titik-titik tadi.
Gambar terlampir.
[tex]~[/tex]
@BeruangGemoy :•
" Life is not a problem to be solved but a reality to be experienced! "
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❥Persamaan Fungsi
f(x) = x² + 5x + 4
[tex] ~ [/tex]
A. Titik potong sumbu x
x² + 5x + 4 = 0
(x + 4)(x + 1) = 0
x = {-4, -1}
Koordinat (-4,0) dan (-1,0)
B. Titik potong sumbu y
maka x = 0
y = x² + 5x + 4
y = 0² + 5(0) + 4
y = 4
Koordinat (0,4)
C. Sumbu simetri
[tex] \tt = \frac{ - b}{2a} [/tex]
[tex] \tt = \frac{ - 5}{2(1)} [/tex]
[tex] \tt = \frac{ - 5}{2} [/tex]
= -2,5
D. Koordinat maksimum
[tex] \tt = \frac{ - b}{2a} , \frac{ - d}{4a} [/tex]
[tex] \tt = ( - \frac{5}{2} , - \frac{ - ( {5}^{2} - 4(1)(4)) }{4(1)} )[/tex]
[tex] \tt = ( - \frac{5}{2} , \frac{ - (25 - 16)}{4} [/tex]
[tex] \tt = ( - \frac{5}{2} , \frac{ - 9}{4} )[/tex]
[tex] \tt = ( - \frac{ 5}{2} , - \frac{9}{4} )[/tex]
= (-2.5 , -2.25)
E. Gambar grafik
Tarik garis pada titik-titik tadi.
Gambar terlampir.
[tex]~[/tex]
@BeruangGemoy :•