Solving Quadratic Problem: y=a(x^2)+bx+c

[Cmunro]

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

 

[Hootenanny]

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)];

\left.\begin{array}{rcr}<br /> 16a + 4b + 12 &amp; = &amp; 0\\<br /> 4b + 2b + 12 &amp; = &amp; 8<br /> \end{array}\right\}<br />

From this you should be able to solve for the coefficients of x (a & b). Can you go from here?

 

[Cmunro]

I can go on from here. Thank you very much!

 

[Hootenanny]

I can go on from here. Thank you very much!

My pleasure