Question

Pls help, thank you

Answer 1

Answer:

• y = -3x² + 7x - 9

Step-by-step explanation:

Given quadratic function:

• y = ax² + bx + c

Points on the graph:

• (-2,-35), (1,-5), (3,- 15)

Substitute values of x and y and solve the system of equations:

• -35 = a(-2)² + b(-2) + c ⇒ -35 = 4a - 2b + c   ⇔ eq 1
• -5 = a(1)² + b(1) + c ⇒ -5 = a + b + c               ⇔ eq 2
• -15 = a(3)² + b(3) + c ⇒ -15 = 9a + 3b + c       ⇔ eq 3

Subtract eq 2 from eq 1:

• -35 - (-5) = 4a - 2b + c - a - b - c
• -30 = 3a - 3b
• b = a + 10                                                         ⇔ eq 4

Subtract eq 2 from eq 3:

• -15 - (-5) = 9a + 3b + c - a - b - c
• -10 = 8a + 2b
• b = -4a - 5                                                       ⇔ eq 5

Compare eq 4 and eq 5, solve for a:

• a + 10 = -4a - 5
• a + 4a = -5 - 10
• 5a = -15
• a = -3

Find the value of b using eq 4:

• b = -3 + 10
• b = 7

Find the value of c using eq 2:

• -5 = -3 + 7 + c
• c = -5 - 4
• c = -9

We now have a, b and c:, the function is:

• y = -3x² + 7x - 9

Answer 2

Answer:

y = -3x² + 7x - 9

Step-by-step explanation:

Given quadratic function:

y = ax² + bx + c

Points on the graph:

(-2,-35), (1,-5), (3,- 15)

Substitute values of x and y and solve the system of equations:

-35 = a(-2)² + b(-2) + c ⇒ -35 = 4a - 2b + c   ⇔ eq 1

-5 = a(1)² + b(1) + c ⇒ -5 = a + b + c               ⇔ eq 2

-15 = a(3)² + b(3) + c ⇒ -15 = 9a + 3b + c       ⇔ eq 3

Subtract eq 2 from eq 1:

-35 - (-5) = 4a - 2b + c - a - b - c

-30 = 3a - 3b

b = a + 10                                                         ⇔ eq 4

Subtract eq 2 from eq 3:

-15 - (-5) = 9a + 3b + c - a - b - c

-10 = 8a + 2b

b = -4a - 5                                                       ⇔ eq 5

Compare eq 4 and eq 5, solve for a:

a + 10 = -4a - 5

a + 4a = -5 - 10

5a = -15

a = -3

Find the value of b using eq 4:

b = -3 + 10

b = 7

Find the value of c using eq 2:

-5 = -3 + 7 + c

c = -5 - 4

c = -9

We now have a, b and c:, the function is:

y = -3x² + 7x - 9