A quadratic relation in the form y = ax² + bx + c and has a y-intercept of (0, 1). The parabola also goes through the points (2, 9) and (-5, 16). Determine the parameters of this quadratic relation (the values of a, b, and c).

value of

a = 1

b = 2

c =1

and equation is

y = x² + 2x + 1

Step-by-step explanation:

Given quadratic equation  y = ax² + bx + c

It has  y-intercept of (0, 1)

thus, point (0,1) will satisfy equation  y = ax² + bx + c.

putting 1 in place of y and 0 in place of x in equation  y = ax² + bx + c we have

1 = a*0² + b*0 + c

c = 1

Thus, equation until now is y = ax² + bx + 1

The parabola also goes through the points (2, 9) and (-5, 16)

point (2, 9) will satisfy equation  y = ax² + bx + 1.

putting 9 in place of y and 0 in place of 2 in equation  y = ax² + bx + 1 we have

9 = a*2² + b*2 + 1

9 = 4a+2b+1

=>4a+2b = 9-1 = 8

dividing both side by 2 we have

4a/2+2b/2  = 8/2

=> 2a + b = 4

b = 4-2a

The parabola also goes through the points (2, 9) and (-5, 16)

point (-5, 16) will satisfy equation  y = ax² + bx + 1.

putting 9 in place of y and 0 in place of 2 in equation  y = ax² + bx + 1 we have

16 = a*(-5)² + b*(-5) + 1

16 = 25a-5b+1

=>25a-5b = 16-1 = 15

dividing both side by 5 we have

25a/5-5b/5 = 15/5

=> 5a - b = 3

second equation is 5a - b = 3

substituting value of b as 4-2a from first equation we have

5a - (4-2a) = 3

=> 5a - 4+2a = 3

=> 7a = 3+4 = 7

=> a = 7/7 = 1

value of b is 4-2a

substituting value of a as 1 in 4 - 2a we have

b = 4 - 2*1 = 4-2 = 2

Thus,

value of

a = 1

b = 2

c =1

and equation is

y = x² + 2x + 1

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: