Question 113829
First let's graph {{{y=8}}}


In order to graph {{{y=8}}}, simply draw a horizontal line through the y=8


{{{ graph( 500, 500, -10, 10, -10, 10, 8) }}}  Graph of  {{{y=8}}}



Now let's move onto the second equation {{{2y=x^2}}}


{{{2y=x^2}}} Start with the given equation



{{{y=x^2/2}}} Divide both sides by 2 to solve for y



Now let's graph {{{y=x^2/2}}} 



In order to do so, let's make a table by plugging in x-values (you get to choose which ones) to find the y-values


<pre>

       x          y
   -5.00000   12.50000
   -4.00000    8.00000
   -3.00000    4.50000
   -2.00000    2.00000
   -1.00000    0.50000
    0.00000    0.00000
    1.00000    0.50000
    2.00000    2.00000
    3.00000    4.50000
    4.00000    8.00000
    5.00000   12.50000

</pre>



Now let's plot these points and connect them to get this graph



{{{ graph( 500, 500, -10, 10, -10, 10, x^2/2) }}} Graph of  {{{y=x^2/2}}} 



Now let's plot the two graphs together:


{{{ graph( 500, 500, -10, 10, -10, 10,8, x^2/2) }}} Graph of {{{y=8}}} (red) {{{y=x^2/2}}} (green)



From the graph, we can see that the two lines intersect at x=-4 and x=4 which means the solutions are (-4,8) or (4,8)






=====================================



Now let's solve algebraically:


Start with the given system

{{{y=8}}}

{{{2y=x^2}}} 



{{{2(8)=x^2}}} Take the 2nd equation and plug in {{{y=8}}}



{{{16=x^2}}} Multiply



*[Tex \LARGE x=\pm 4] Take the square root of both sides



So we then get {{{x=-4}}} or {{{x=4}}}


So this means our solutions are (-4,8) or (4,8) (remember y is always 8 since the first equation is {{{y=8}}})