Find the vertex of quadratic function y=x^2-2x-15 Plis helppp,,,,

[tex]Fungsi \:Kuadrat[/tex]

fungsi kuadrat y = x² - 2x - 15
memiliki nilai a = 1 , b = -2, c = -15

vertex (koordinat titik balik) terletak pada sumbu simetri xs
xs = -(b/2a)
xs = -(-2/2)
xs = 1

nilai baliknya adalah y(xs) = y(1)
y(1) = 1² - 2(1) - 15
y = -16

jadi, koordinat titik baliknya adalah (1 , -16)

semoga jelas dan membantu


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in english

a quadratic function y = x² - 2x - 15
has a = 1 , b = -2, c = -15

the vertex lies on axis of symmetry xs
xs = -(b/2a)
xs = -(-2/2)
xs = 1

and the optimum value is
y(xs) = y(1)
y(1) = 1² - 2(1) - 15
y = -16

hence, the vertex is (1 , -16)

Mapel Matematika
Bab Koordinat

Vertex = Koordinat Titik Balik

y = x^2 - 2x - 15

Ordinat
= -b/2a
= -(-2)/2(1)
= 1

Absis
= 1^2 - 2(1) - 15
= -16

Koordinat TB (1, -16)