SOLUTION: Could someone help me with the following problem? Problem : A quadratic funtion has its vertex at the point of (0,-5). The function passes throuhg the point (-5,2). When w

Algebra ->  Equations -> SOLUTION: Could someone help me with the following problem? Problem : A quadratic funtion has its vertex at the point of (0,-5). The function passes throuhg the point (-5,2). When w      Log On


   

Question 189678: Could someone help me with the following problem?
Problem : A quadratic funtion has its vertex at the point of (0,-5). The function passes throuhg the point (-5,2). When written in vertex form the function is f(x)=a(x-h)^2+k. Where a = ? , h=?, and k=?
Ok, I figured out what h and k are ( h = 0 , and k = -5) but I dont know how to get a.. could someone help me.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=a%28x-h%29%5E2%2Bk Start with the general vertex form.


f%28x%29=a%28x-0%29%5E2%2B%28-5%29 Plug in h=0 and k=-5


f%28x%29=ax%5E2-5 Simplify

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2=a%28-5%29%5E2-5 Plug in x=-5 and f%28x%29=2


Note: since the graph passes through the point (-5,2), this means that if x=-5, then f%28x%29=2 (f(x) is really "y")


2=a%2825%29-5 Square -5 to get 25


2=25a-5 Rearrange the terms


2%2B5=25a Add 5 to both sides.


7=25a Combine like terms.


7%2F25=a Divide both sides by 25 to isolate "a".


a=7%2F25 Rearrange the equation


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f%28x%29=ax%5E2-5 Go back to the previous function


f%28x%29=%287%2F25%29x%5E2-5 Plug in a=7%2F25




So the function with a vertex of (0,-5) and passes through (-5,2) is f%28x%29=%287%2F25%29x%5E2-5