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anonymous
 one year ago
For the given quadratic equation convert into vertex form, find the vertex, and find the value for x = 6. y= 2x^2 + 2x +2?
anonymous
 one year ago
For the given quadratic equation convert into vertex form, find the vertex, and find the value for x = 6. y= 2x^2 + 2x +2?

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if we make this substitution: x=6 into your quadratic equation, we get: \[\Large \begin{gathered} y =  2 \times {6^2} + 2 \times 6 + 2 = \hfill \\ \hfill \\ =  2 \times 36 + 12 + 2 = ...? \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that's the value of x, correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2that's the value of y, when x=6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how do we find the value of x and find the vertex?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2our task was to find the value of y, when x=6

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the general formula of a parabola is: \[\Large y = a{x^2} + bx + c\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So there is nothing else to the problem it's done that simple lol

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to find the coordinates of the vertex, and the vertex form of the equation of your parabola

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, as I wrote before, the general equation of a parabola, whose axis is parallel to y axis, is: \[\Large y = a{x^2} + bx + c\] now, please by comparison with the equationof your parabola, what are the coefficients a, b, and c?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: dw:1435870805196:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now the xcoordinate of the vertex is: \[\Large x =  \frac{b}{{2a}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please substitute your coefficients into taht formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0X=\[\frac{ 2b }{ 2a}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: \[\Large x =  \frac{b}{{2a}} =  \frac{2}{{2 \times \left( {  2} \right)}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large x =  \frac{b}{{2a}} =  \frac{2}{{2 \times \left( {  2} \right)}} =  \frac{2}{{  4}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I Oh ok i thought it was like a trick next step lol

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next, the ycoordinate of the vertex is given by the subsequent formula: \[\Large y =  \frac{{{b^2}  4ac}}{{4a}} = ...?\] as before, please substitute your coefficients into that formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 2^{2}4ac }{ 4a}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: \[\Large y =  \frac{{{b^2}  4ac}}{{4a}} =  \frac{{{2^2}  \left\{ {4 \times \left( {  2} \right) \times 2} \right\}}}{{4 \times \left( {  2} \right)}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 20 }{ 8}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so its \[2\frac{ 1 }{ 2 }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! so the vertex V of our parabola is this point: \[\Large V = \left( {\frac{1}{2},\frac{5}{2}} \right)\] since: \[\Large \frac{{20}}{8} = \frac{5}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So my answers are 58 for x inputting into the y and my vertex is (1/2) (5/2)? correct

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2finally, we have to write the vertex form of the equation of your parabola

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much for your patience

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2first step: we can factor out 2 at the right side of your equation, so we get: \[\Large y =  2\left( {{x^2}  x  1} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do we input abc or leave it like that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to continue with the second step

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here is the second step: \[\Large {x^2}  x  1 = {x^2}  x + \frac{1}{4}  \frac{1}{4}  1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I added and subtracted 1/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so wouldnt it just x^2x1

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! I cahnge the form of x^2x1 using the rules of algebra

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2nevertheless the quantity is the same, I change only its algebraic shape

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes that quantity will be the same, I change its form only

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next I can simplify that expression above as below: \[\Large \begin{gathered} {x^2}  x  1 = {x^2}  x + \frac{1}{4}  \frac{1}{4}  1 = \hfill \\ \hfill \\ = {x^2}  x + \frac{1}{4}  \frac{5}{4} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, the quantity: \[\large {x^2}  x + \frac{1}{4}\] is a perfect square, namely it is the square of which binomial?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: please compute this: \[\Large {\left( {x  \frac{1}{2}} \right)^2} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435879978188:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we can write this: \[\large \begin{gathered} {x^2}  x  1 = {x^2}  x + \frac{1}{4}  \frac{1}{4}  1 = \hfill \\ \hfill \\ = {x^2}  x + \frac{1}{4}  \frac{5}{4} = {\left( {x  \frac{1}{2}} \right)^2}  \frac{5}{4} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2namely: \[\large {x^2}  x  1 = {\left( {x  \frac{1}{2}} \right)^2}  \frac{5}{4}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2then I substitute that new expression, into the subsequent formula: \[\Large y =  2\left( {{x^2}  x  1} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and I get: \[\Large y =  2\left\{ {{{\left( {x  \frac{1}{2}} \right)}^2}  \frac{5}{4}} \right\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=2\left( x\frac{ 1 }{ 2 } \right)^{2}+\frac{ 5 }{ 2 }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2congratulations!! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the vertex is 1/2 and 5/2, that's are equation above and 58 is the input?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay thank you so much!
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