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.

Guest Aug 15, 2019

#1**+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?

!

asinus Aug 15, 2019

#2**+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

CPhill Aug 15, 2019