Tolong bantu jawab kk

Tolong bantu jawab kk

Jawaban Terkonfirmasi

f(x) = (t – 3)x³ + 2tx + 5 ——> a = (t – 3) ; b = 2t ; c = 5
nilai max = $frac{D}{-4a}$
9 = $frac{ b^{2} - 4.a.c}{-4a}$
9 = $frac{ (2t)^{2}- 4.(t - 3).5 }{-4(t - 3)}$
-36(t – 3) = 4t² – 20t + 60
-36t + 108 = 4t² – 20t + 60
0 = 4t² + 36t – 20t – 108 + 60
0 = 4t² + 16t – 48
0 = t² + 4t – 12
0 = (t + 6)(t – 2)
t + 6 = 0 atau t – 2 = 0
t = -6 atau t = 2
untuk t = 6
f(x) = (-6 – 3)x² + 2(-6)x + 5
f(x) = -9x² – 12x + 5 ——-> sumbu simetri = $- frac{b}{2a}$
   = - $frac{-12}{2. -9}$
   = - $frac{-12}{-18}$
   = - $frac{2}{3}$
untuk t = 2
f(x) = (2 – 3)x² + 2(2)x + 5
f(x) = – x² + 4x + 5 ——-> sumbu simetri = - $frac{b}{2a}$
= - $frac{4}{2. -1}$
= - $frac{4}{-2}$
= 2
jadi,sumbu simetrinya adalah - $frac{2}{3}$ dan 2