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

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

===============================================

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

User Mikemil
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