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avatar+1347 

Help, I can't get this

 

There exist constants a, h  and k such that

3x^2 + 12x + 4 - 2x^2 + 16x - 25 = a(x - h)^2 + k
for all real numbers  Enter the ordered triple (a,h,k).

 Apr 25, 2023
 #1
avatar+2667 
+1

Simplify the left-hand side to \(x^2 + 28x - 21\).

 

Now, notice that the right-hand side is just vertex form. This means that a is the coefficient of the \(x^2\) term (in this case it's 1) and that \((h, k)\) is the vertex. 

 

The vertex is located at \(x = -{b \over 2a} = -14\). Substituting this in gives us \((-14)^2 + 28(-14) - 21 = -217\). So, the vertex is \((-14, -217)\).

 

Thus, the ordered triple is just \(\color{brown}\boxed{(1, -14, -217)}\)

 Apr 25, 2023

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