Question

Use quadratic regression to find the equation for the parabola going through these 3 points (-5, 13) (0, 8) and (2, 48)

Answer 1

Answer:
`\text{y} = 3\text{x}^2+14\text{x}+8`

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Step-by-step explanation

There are many approaches we can take here, so feel free to explore other options. The method I'll use is substitution more or less.

The quadratic template is
`\text{y} = a\text{x}^2 + b\text{x}+c`

If we were to plug the coordinates of (x,y) = (-5,13) then

`\text{y} = a\text{x}^2 + b\text{x}+c\\\\13 = a(-5)^2 + b(-5)+c\\\\25a-5b+c = 13`

Repeat similar steps for the point (0,8) and you should find that c = 8.

Let's plug that into the equation we just found.

`25a-5b+c = 13\\\\25a-5b+8 = 13\\\\25a-5b = 13-8\\\\25a-5b = 5\\\\5(5a-b) = 5\\\\5a-b = 5/5\\\\5a-b = 1\\\\`

Now solve for b

`5a-b = 1\\\\5a = 1+b\\\\b = 5a-1`

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Let's plug (x,y) = (2,48) into the template. We'll use c = 8 and b = 5a-1 as well.

`\text{y} = a\text{x}^2+b\text{x}+c\\\\48 = a(2)^2+b(2)+c\\\\48 = 4a+2b+c\\\\48 = 4a+2(5a-1)+8\\\\48 = 4a+10a-2+8\\\\48 = 14a+6\\\\14a = 48-6\\\\14a = 42\\\\a = 42/14\\\\a = 3`

Use this to determine b.

`b = 5a-1\\\\b = 5*3-1\\\\b = 14`

In summary:

a = 3, b = 14, c = 8

The template
`\text{y} = a\text{x}^2+b\text{x}+c`

updates to the final answer
`\text{y} = 3\text{x}^2+14\text{x}+8`