Question #91935
Find the X value for the vertex : (show all work)
1. f(x)=x^2+4x+6
2.f(x)=2(x-4)^2+5
3.f(x)=-2(x+4)^2+3
4.f(x)=-12x^2+12x+3
1
Expert's answer
2019-07-25T14:02:52-0400
f(x)=x2+4x+6f(x)=x^2+4x+6

Solution:


f(x)=(x+2)2+2f(x)=(x+2)^2+2



(x+2)2=0;x+2=0;x=2;(x+2)^2=0; x+2=0;x=-2;x=2=>0+2=y; y=2x=-2 => 0+2=y;\space y=2

Answer: (2,2);(-2,2);

2.


f(x)=2(x4)2+5f(x)=2(x-4)^2+5

Solution:

2(x4)2=0; x4=0; x=42(x-4)^2=0;\space x-4=0;\space x=4x=4=>0+5=y; y=5x=4 => 0+5=y;\space y=5

Answer: (4,5);(4,5);

3.

f(x)=2(x+4)2+3f(x)=-2(x+4)^2+3

Solution:

2(x+4)2=0; x+4=0; x=4-2(x+4)^2=0;\space x+4=0;\space x=-4


x=4=>0+3=y; y=3x=-4 => 0+3=y;\space y=3

Answer:(4,3);(-4,3);

4.

f(x)=12x2+12x+3f(x)=-12x^2+12x+3

Solution:

f(x)=3(2x1)2+6;f(x)=-3(2x-1)^2+6;

3(2x1)2=0; 2x1=0; x=12;-3(2x-1)^2=0; \space2x-1=0;\space x=\frac{1}{2};

x=12=> 0+6=y; y=6x=\frac{1}{2}=>\space 0+6=y;\space y=6

Answer:(12,6);(\frac{1}{2},6);


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