Jawab:

Penjelasan dengan langkah-langkah:

dikatakan definit positif saat D < 0 & a > 0

dikatakan definit negatif saat D < 0 & a < 0

1. (p+1)x² -2px + (p-4), definit negatif saat

a < 0

(p+1) < 0

p < -1 ...(hp1)

D = b² - 4.a.c < 0

(-2p)² -4.(p+1).(p-4) < 0

4p² -4(p²-3p-4) < 0

4p² -4p²+12p+16 < 0

12p + 16 < 0

p < -16/12

p < -1 ...(hp2)

HP = hp1 ∩ hp2 = {p<-1}

2. (p+1)x² +(p+2)x - (p-1), definit positif saat

a > 0

(p+1) > 0

p > -1 ...(hp1)

D = b² - 4.a.c < 0

(p+2)² -4.(p+1).-(p-1) < 0

(p²+4p+4) +4(p²-1) < 0

5p²+4p < 0

p(5p+4) < 0

-4/5 < p < 0 ...(hp2)

HP = hp1 ∩ hp2 = {-4/5 < p < 0}

3.y = x² + ax + b.

titik min = (Xmin, Ymin) = (1,2)

Xmin = -b/2a

1 = -a/2

a = -2

masukkan nilai a dan nilai (Xmin, Ymin) = (1,2) ke persamaan.

y = x² + ax + b

2 = 1² + (-2)1 + b

2 = 1-2 + b

b = 3

a,b = {-2,3}

4.y = 2x² -4x + 7

Xmin = -b/2a

Xmin = -(-4)/2.2 = 1

Ymin = f(Xmin) =  2.1² -4.1 + 7

Ymin = 2-4+7 = 5

Titik puncak = {Xmin, Ymin} = {1,5}

5.y = 2(x-3)²-15

y = 2(x² -6x + 9) -15

y = 2x² -12x + 18 -15

y = 2x² -12x + 3

Sumbu simetri saat nilai D = 0, dengan kata lain sumbu simetri bernilai sama dengan Xmin.

Xmin = -b/2a

Xmin = -(-12)/2.2 = 3

Sumbu simetri x = 3.