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1. Let f(x) be deinied as f(x)=x^2+8x+12. If you were take f(x) and translate the vertex down 3 units, what "a" value would you need to use in order to have the same x intercepts for the new etqion as f(x)?

 

2. Given the standard format of a quadractiv is y=ax^2+bx+c, use the complete square process to derive th quadratic equations.

 Aug 15, 2019
 #1
avatar+8792 
+1

1. Let f(x) be deinied as f(x)=x^2+8x+12. If you were take f(x) and translate the vertex down 3 units, what "a" value would you need to use in order to have the same x intercepts for the new etqion as f(x)?

 

1. That means:
What is the equation of function of a parabola with the zeroes -6 and -2 and the vertex S (-4; -7)?
How do you design these?
Sequel follows.

What does "new etqion" mean?

laugh  !

 Aug 15, 2019
edited by asinus  Aug 15, 2019
 #2
avatar+106515 
+1

. Let f(x) be deinied as f(x)=x^2+8x+12. If you were take f(x) and translate the vertex down 3 units, what "a" value would you need to use in order to have the same x intercepts for the new equation as f(x)?

 

The x intercepts   for   f(x)  are   (x + 6) (x + 2) = 0    ⇒      x = -6   and x  = -2

 

The x coordinate of the  vertex is     -8/ (2 (1) ) =   -4

 

And the y  coordinate of the vertex  is  (-4)^2 + 8(-4) + 12 =  16 - 32 + 12 =  -4

 

So.....the  y  coordinate for the  new  vertex  is    -7

 

And the  x  coordinate  of the  vertex is   -4

 

So  the  point  (-4, -7) is on the new graph.....therefore

 

a(-6 + 4)^2  -  7    =  0

 

a (-2)^2 - 7  =0

 

4a  - 7   =0

 

4a  = 7

 

a  = 7/4

 

See the graph here :    https://www.desmos.com/calculator/av5zvze5hk

 

 

cool cool cool

 Aug 15, 2019

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