anonymous
  • anonymous
Create an equation. Use the graph below to create the equation of the rainbow parabola.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@rishavraj
anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/557704cee4b03f31cab31848-maryslater-1433863506866-06_05_06_012.gif
rishavraj
  • rishavraj
so wht do u thiink wht r the roots of the equation???

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anonymous
  • anonymous
um -6 and 6
rishavraj
  • rishavraj
cool......xD means u don't need any help :))
anonymous
  • anonymous
lol! I so I got -x^2+36 ??
jdoe0001
  • jdoe0001
well... notice whre the vertex is at the vertex is at (0, 36) well.. actaully that's quite close, one sec
anonymous
  • anonymous
and there's a 2nd part that I'm not understading
anonymous
  • anonymous
Create a table of values for a linear function. A drone is in the distance, flying upward in a straight line. It intersects the rainbow at two points. Choose the points where your drone intersects the parabola and create a table of at least four values for the function. Remember to include the two points of intersection in your table.
jdoe0001
  • jdoe0001
\(\bf y=a(x-{\color{brown}{ h}})^2+{\color{blue}{ k}} \qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\) the parabola is upside-down, meaning the "a" is negative, yes thus \(\bf y=a(x-{\color{brown}{ h}})^2+{\color{blue}{ k}} \qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}}) \\ \quad \\ y=-a(x-0)^2+36\implies y=-ax^2+36\) we know the roots, thus, when y =0 , x = 6 or -6, let's use 6 so 6, 0 then \(\bf y=-ax^2+36\implies \cfrac{y-36}{-x^2}=a\implies \cfrac{0-36}{-6^2}=a\)
anonymous
  • anonymous
Yes, so it would be -x^2+36 :)
jdoe0001
  • jdoe0001
ohh yeah, it would, anyhow, kudos :)
jdoe0001
  • jdoe0001
lemme check the other bit
anonymous
  • anonymous
Thanks:) Because I have no idea what to do
jdoe0001
  • jdoe0001
well, is a system of equations lemme see if hmmm lemme graph it
anonymous
  • anonymous
Ok.
jdoe0001
  • jdoe0001
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIteF4yKzM2IiwiY29sb3IiOiIjRTMxMjEyIn0seyJ0eXBlIjowLCJlcSI6IigxLzMpeCsyNSIsImNvbG9yIjoiIzIyMUJFMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi0xMy40MjQ4MzUyMDUwNzgxMTgiLCIxMS45NjU3ODk3OTQ5MjE4NjgiLCIyMS4xMDg2MjczMTkzMzU5MzgiLCIzNi43MzM2MjczMTkzMzU5MyJdfV0- you have the "rainbow" parabola and then you're asked to make a LINEar function you're expected to use a table of values for the linear one and to include at least, four values, more is fine so if you use 10 values is ok as well, but not less than 4 values for the table notice the graph.... the LINEar equation graph, crosses the parabola at two points
anonymous
  • anonymous
how did you get 1/3x+25?
jdoe0001
  • jdoe0001
made up :), you're expected to make one up, thus
jdoe0001
  • jdoe0001
I simply made sure the line was secant to the parabola
jdoe0001
  • jdoe0001
but you can cross the parabola by other lines as well you could simply pick two points, to make a line one on the left, one on the right make sure they "cut" the parabola through, so they touch it at two points and with two points you can get the equation of that line from there you can get more values for the table
anonymous
  • anonymous
ok, so what would the points be?
jdoe0001
  • jdoe0001
well.. let's use the 1/3x + 25 one sec
anonymous
  • anonymous
why couldn't we use like 9/5x+20 or 2x+27? And okay, ill wait:)
jdoe0001
  • jdoe0001
either one will work, so long they "cut through" the parabola
anonymous
  • anonymous
okay:) So 9/5x+20 would cut through right?
jdoe0001
  • jdoe0001
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIteF4yKzM2IiwiY29sb3IiOiIjRTMxMjEyIn0seyJ0eXBlIjowLCJlcSI6IigxLzMpeCsyNSIsImNvbG9yIjoiIzIyMUJFMCJ9LHsidHlwZSI6MCwiZXEiOiI5LzV4KzIwIiwiY29sb3IiOiIjMTNFMzBDIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTI2LjMwNjMwNDkzMTY0MDYwNyIsIjIzLjI4NDc1OTUyMTQ4NDM1NyIsIjcuMzM0MjEzMjU2ODM1OTQ2IiwiMzcuODUxNzkxMzgxODM1OTMiXX1d
anonymous
  • anonymous
Thank you:)! And to find the points, we could just plug in numbers? or does it have to be specific
jdoe0001
  • jdoe0001
\(\begin{array}{rrllll} x&y\\ \\\hline\\ -6&\frac{1}{3}\cdot -6+36\\ -4&\frac{1}{3}\cdot -4+36\\ -2&\frac{1}{3}\cdot -2+36\\ 0&\frac{1}{3}\cdot 0+36\\ 2&\frac{1}{3}\cdot 2+36\\ 4&\frac{1}{3}\cdot 4+36\\ 6&\frac{1}{3}\cdot 6+36\\ \end{array}\) there, that'd a table for the blue one
jdoe0001
  • jdoe0001
yw
anonymous
  • anonymous
Yayyy! :) Lol I finally got it:) Thanks so muchhh
jdoe0001
  • jdoe0001
np

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