+895 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 \(y=(x+4)^2+7\).
+129852 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
