Question 166497
# 1



{{{y=2(x+3)^2-5}}} Start with the given equation



{{{y=2(x-(-3))^2-5}}} Rewrite {{{x+3}}} as {{{x-(-3)}}}



Now the equation above is in vertex form {{{y=a(x-h)^2+k}}} where (h,k) is the vertex and 'a' determines which direction it opens up. In this case, {{{a=2}}}, {{{h=-3}}} and {{{k=-5}}}. 



So because {{{a>0}}}, this tells us that the parabola opens up. Also, since {{{h=-3}}} and {{{k=-5}}}, the vertex is (-3,-5). Finally, the axis of symmetry is simply the x-coordinate of the vertex. So the axis of symmetry is {{{x=-3}}}



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Answer:


Direction of opening:  up 

Vertex: (-3,-5)

Axis of Symmetry: {{{x=-3}}}



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# 2




{{{-4x^2+8x-1}}} Start with the right side of the given equation.



{{{-4(x^2-2x+1/4)}}} Factor out the {{{x^2}}} coefficient {{{-4}}}. This step is very important: the {{{x^2}}} coefficient <font size=4><b>MUST</b></font> be equal to 1.



Take half of the {{{x}}} coefficient {{{-2}}} to get {{{-1}}}. In other words, {{{(1/2)(-2)=-1}}}.



Now square {{{-1}}} to get {{{1}}}. In other words, {{{(-1)^2=(-1)(-1)=1}}}



{{{-4(x^2-2x+highlight(1-1)+1/4)}}} Now add <font size=4><b>and</b></font> subtract {{{1}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{1-1=0}}}. So the expression is not changed.



{{{-4((x^2-2x+1)-1+1/4)}}} Group the first three terms.



{{{-4((x-1)^2-1+1/4)}}} Factor {{{x^2-2x+1}}} to get {{{(x-1)^2}}}.



{{{-4((x-1)^2-3/4)}}} Combine like terms.



{{{-4(x-1)^2-4(-3/4)}}} Distribute.



{{{-4(x-1)^2+3}}} Multiply.



So after completing the square, {{{-4x^2+8x-1}}} transforms to {{{-4(x-1)^2+3}}}. So {{{-4x^2+8x-1=-4(x-1)^2+3}}}.


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Answer:


So the answer is {{{y=-4(x-1)^2+3}}} which is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=-4}}}, {{{h=-1}}}, and {{{k=3}}}