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anonymous
 one year ago
Write the quadratic function in the form f(x)=a−(xh)^ 2+k . Then, give the vertex of its graph.f(x)=2x^2+8x5
anonymous
 one year ago
Write the quadratic function in the form f(x)=a−(xh)^ 2+k . Then, give the vertex of its graph.f(x)=2x^2+8x5

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0\(f(x) = ax^2+bx+c\) is the same thing as \(f(x) =a(x+\dfrac{b}{2a})^2+ca(\dfrac{b}{2a})^2 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i just plug it in?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isn't the x point for the vertex b/2a?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0this is not that hard it is \[f(x)=2x^2+8x5\]right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0you want to write it as \[f(x)=a(xh)^2+k\] all you need is \(h\) and \(k\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0since evidently \(a=2\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0to find \(h\) compute \(\frac{b}{2a}\) which, in your case, is \(\frac{8}{2\times (2)}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it 8/4x? and simplify so its 2x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh opps i mistaken x and a variable

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0@Kimes yes just plug things :)
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