+0  
 
0
211
2
avatar+3444 

Help.

NotSoSmart  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 \(y=(x+4)^2+7\).

AdamTaurus  Nov 2, 2017
 #2
avatar+89876 
+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

CPhill  Nov 2, 2017

15 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.