Question

Use the quadratic regression to find the equation for the parabola going through these 3 points (-5, 161) (-2, 29) (6, 205)

Answer 1

Answer:

The equation of the parabola is y = 6$x^{2}$-2b+1

Step-by-step explanation:

First we write a quadratic equation of parabola as y = a$x^{2}$+bx+c

Now it is given in the question that parabola passes through three points (-5,161),(-2,29),(6,205).

To find the equation we have to get the value of a,b & c.

Now we form three equations to find the value of a,b,c.

We put the value (-5,161) in the equation

161 = a×25+b×(-5)+c

161 = 25a-5b+c----------(1)

Now we put (-2,29) in the equation

29 = a×4+b×(-2)+c

29 = 4a-2b+c---------(2)

Again we put (6,205) in the equation

205 = a×36+6b+c------------(3)

Now we subtract equation (1) from (2)

161-29 = 25a-4a-5b-(-2b)+c-c

132 = 21a-5b+2b

132 = 21a-3b

132 = 3(7a-b)

44 = 7a-b---------(4)

Now we subtract equation (2) from (3)

205-29 = 36a-4a+6b-(-2b)+c-c

176 = 32a+6b+2b

176 = 32a+8b

176 = 8(4a+b)

22 = 4a+b------------(5)

Now we add equation (5) and (4)

44+22 = 7a+4a-b+b

66 = 11a

a = 6

Now we put the value a in equation (4)

44 = 7×6-b

44 = 42-b

44-42 = -b

2 = (-b)

b = (-2)

Now we put the value of a and b in equation number (2)

29 = 4×6-2(-2)+c

29 = 24+4+c

29 = 28+c

c = 29-28

c =1

Finally we put the value of a,b,c in the quadratic equation

y = 6$x^{2}$+(-2)x+1

y = 6$x^{2}$-2+1

This is the final answer.