Question 37236
<pre><font size = 5><b>Create the quadratic equation in the form 
y = ax² + bx + c using the point (-1,7) as one
point and the point (10,-8) as the vertex. 
Enter a,b,c values as common fractions in 
reduced form. 

We start out with the standard form:

y = a(x - h)² + k where the vertex is 
(h, k) = (10, -8).  Substituting:

y = a(x - 10)² - 8

Now this must go through the point (-1, 7), 
so substitute -1 for x and 7 for y, and 
solve for a:

           7 = a(-1 - 10)² - 8

           7 = a(-11)² - 8

           7 = a(121) - 8

           15 = 121a

       15/121 = a

Now y = a(x - 10)² - 8 becomes

    y = 15/121(x - 10)² - 8

Clear of fractions temporarily by multiplying 
both sides by 121:

   121y = 15(x - 10)² - 968

   121y = 15(x² - 20x + 100) - 968

   121y = 15x² - 300x + 1500 - 968

   121y = 15x² - 300x + 532

Divide through by 121

      y = (15/121)x² - (300/121)x + 532/121

a = 15/121, b = -300/121, c = 532/121

Edwin
AnlytcPhil@aol.com</pre>