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avatar+4690 

Help.

 Nov 2, 2017
 #1
avatar+771 
+2

Since the standard form of a quadratic is y=a(x-a)2+b, we can plug in the vertex.

\(y=a(x+4)^2+7\)

To find a, plug in the point it passes through.

\(8=a(-3+4)^2+7\)

Simplify.

\(8=a(1)+7\)

\(8=a+7\)

\(1=a\)

 

So, the formula is \(y=(x+4)^2+7\).

 Nov 2, 2017
 #2
avatar+96189 
+2

Good job, AT

 

Here's the second one

 

y = -2x^2  + 32x - 12

 

The x coordinate of the vertex is    -32 / [ 2 ( -2 ) ]  =  -32 / -4  = 8

 

Since the first term is negative, this parabola turns downward, so we have a max at

 

-2 * 8^2  + 32 (8) - 12  =     -128 + 256 - 12 = 116

 

So...no y value can be > 116..so the range is   y ≤ 116 

 

 

 

cool cool cool

 Nov 2, 2017

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