Question

How do I find the equation of a curve y=ax^2+bx+c that passes through (2,5), (3,12) and (-1,-4)?

Answer 1

Step-by-step explanation:

How do I find the equation of a curve y=ax^2+bx+c that passes through (1,8), (-1,2) and (2,14)?

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How do I find the equation of a curve y=ax^2+bx+c that passes through (1,8), (-1, 2) and (2,14)?

x=1,y=8

means that

8=a(1)2+b(1)+c.

Label this as equation (1):

8=a+b+c……(1).

x=−1,y=2

means that

2=a(−1)2+b(−1)+c.

This simplifies to equation (2):

2=a−b+c……(2)

x=2,y=14

means that

14=a(2)2+b(2)+c.

This is simplified to get equation (3):

14=4a+2b+c……(3)

Add equation (1) and equation (2) to eliminate b and get

2a+2c=10.

Divide this by 2 and obtain the next equation, that is, equation (4):

a+c=5……(4).

Add 2 times equation (2) to equation (3) to eliminate b again and get

6a+3c=18.

Divide this through by 3 and get equation (5):

2a+c=6……(5)

Subtract equation (4) from equation (5) to eliminate c and get

a=1.

Then substitute this into equation (4) and get

1+c=5

so that

c=4.

Go back to equation (1) and replace a and c by these results and get that

1+b+4=8,

from which

b=3.

∴y=x2+3x+4