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The parabola y=ax^2+bx+c has vertex (p,p) and y-intercept (0,p) where \(p\ne 0\). What is b?

 Feb 20, 2019

Best Answer 

 #1
avatar+5038 
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\(\text{I don't think this works}\\ \text{The vertex at }(p,p) \text{ means we can write the parabola as }\\ y = a(x-p)^2 + p\\ \text{and at }x=0\\ y = a(-p)^2 + p = p\\ ap^2 = 0 \Rightarrow a = 0 \vee p = 0\\ \text{we're told }p \neq 0 \text{ so }a=0\\ \text{if }a=0 \text{ we don't have a parabola}\)

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 Feb 20, 2019
 #1
avatar+5038 
+1
Best Answer

\(\text{I don't think this works}\\ \text{The vertex at }(p,p) \text{ means we can write the parabola as }\\ y = a(x-p)^2 + p\\ \text{and at }x=0\\ y = a(-p)^2 + p = p\\ ap^2 = 0 \Rightarrow a = 0 \vee p = 0\\ \text{we're told }p \neq 0 \text{ so }a=0\\ \text{if }a=0 \text{ we don't have a parabola}\)

Rom Feb 20, 2019

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