anonymous
  • anonymous
y=2x^2-8x+7 i have to find the axis of symmetry the coordinates of the vertex and tell whether the vertex is a maximum or a minimum
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
-b/2a ia axis of symmetry --8/2.2 8/4 = 2
amistre64
  • amistre64
ia is stupidity for "is" ...
anonymous
  • anonymous
haha thank you

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anonymous
  • anonymous
\[y = 2x^2 - 8x + 7\] \[ = 2(x^2 - 4x) + 7\] \[= 2(x^2 -4x + 4) -8 + 7\] \[\implies y +1 = 2(x-2)^2\]
anonymous
  • anonymous
So we have a parabola which is shifted 1 unit up in y and 2 units forward in x and grows twice as fast as the standard parabola.
anonymous
  • anonymous
Err shifted 1 unit down in y, sorry.
anonymous
  • anonymous
oh ok u got me confused 4 a sec
anonymous
  • anonymous
what part of the problem are u helping with
anonymous
  • anonymous
Finding the vertex, and the axis of symmetry.
anonymous
  • anonymous
Actually this form of the function answers all of those questions.
anonymous
  • anonymous
i did not learn it that way can u help me another way by chance
anonymous
  • anonymous
Since the coefficient of x is positive, the graph is opening upward. Therefore the vertex will be a min. The location of the vertex can be found from the shift in x and y of the function from (0,0). In this case we shifted down 1 unit and 2 units to the right. \[y-y_0 = m(x-x_0)^2\] Is the standard form for a parabola. Where \(x_0,y_0\) are the x and y points of your vertex. And m is the scaling and directional coefficient. So we have \(y - (-1) = 2(x - 2)^2\) which tells us the vertex is at (2,-1) And since the graph is opening upward the line x = 2 will be the line of symmetry (the line going up/down that goes through the vertex)
anonymous
  • anonymous
thank you soooooo much

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