SOLUTION: How would I go about writing the quadratic function, in (standard) form, which has a vertex of (-6, 2) and passes through (5, 12)? The standard form is y= a(x-h)^2+k , so i woul

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: How would I go about writing the quadratic function, in (standard) form, which has a vertex of (-6, 2) and passes through (5, 12)? The standard form is y= a(x-h)^2+k , so i woul      Log On


   

Question 151404: How would I go about writing the quadratic function, in (standard) form, which has a vertex of (-6, 2) and passes through (5, 12)?
The standard form is y= a(x-h)^2+k , so i would put -6 as h and 2 as k but from there I am stuck.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
y=+a%28x-h%29%5E2%2Bk Start with the given equation.


y=+a%28x-%28-6%29%29%5E2%2B2 Plug in h=-6 and k=2


y=+a%28x%2B6%29%5E2%2B2 Rewrite x-%28-6%29 as x%2B6


Since the quadratic passes through (5, 12), this means that if you plug in x=5, you will get y=12. So we can plug in x=5 and y=12 to solve for "a"

12=+a%285%2B6%29%5E2%2B2 Plug in x=5 and y=12


12=+a%2811%29%5E2%2B2 Add


12=+a%28121%29%2B2 Square 11 to get 121


12=+121a%2B2 Rearrange the terms


10=+121a Subtract 2 from both sides.


10%2F121=+a Divide both sides by 121.


So the value of "a" is a=10%2F121



This means that the equation is


y=+%2810%2F121%29%28x%2B6%29%5E2%2B2