Answer to Question #91935 in Algebra for Karla Carreras

Question

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

Answer 1

$$f(x)=x^2+4x+6$$

Solution:

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

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

Answer: $(-2,2);$

2.

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

Solution:

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

Answer: $(4,5);$

3.

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

Solution:

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

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

Answer:$(-4,3);$

4.

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

Solution:

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

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

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

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