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A quadratic problem

  • Thread starter Cmunro
  • Start date
34
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I have been given a root of 4, a y-intercept of 12 and a known point of (2,8). I have then been asked to write the equation in the form of y=a(x^2)+bx+c

I am assuming the relevant equations are: y=a(x-"alpha")(x-"beta")

Ok, so I know that c=12, but I can't see how I draw the a and b part of the general form from the facts that I have been given. I tried the y=a(x-"alpha")(x-"beta") equation in hopes that I could then expand it out to general form. In this equation alpha=4, however when I substitute y and x with the point (2,8) I am still left with 2 variables: beta and a.

What am I missing?

Thanks, Cat
 

Answers and Replies

Hootenanny
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You have done everything correctly, the next step is to set up a system of equations using the data you have been given. You know that when you substitute the root in for x, the result must be zero. When you substitute 2 in for x, the result must be 8 [from your point (2,8)];

[tex]\left.\begin{array}{rcr}
16a + 4b + 12 & = & 0\\
4b + 2b + 12 & = & 8
\end{array}\right\}
[/tex]

From this you should be able to solve for the coefficients of x (a & b). Can you go from here?
 
Last edited:
34
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I can go on from here. Thank you very much!
 
Hootenanny
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Science Advisor
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9,598
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