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Persamaan sumbu simetri f(x)= -x²+10x-21

persamaan sumbu simetri f(x)= -x²+10x-21

Jawaban soal ini adalah 𝑥 = 5

Ingat!
Bentuk umum fungsi kuadrat
y = f (𝑥) = a𝑥² + b𝑥+ c

Rumus sumbu simetri
𝑥 = -b/2a

Pembahasan
f(x)= -x²+10x-21

a = – 1
b = 10
c = – 21

Rumus sumbu simetri
𝑥 = -b/2a
= – 10/[2 (-1)] = -10 /(-2)
= 5

Sehingga dapat disimpulkan bahwa, sumbu simetrinya adalah 𝑥 = 5.

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