Point Cis on line segment BD. Given BD = 2x, CD = 8, and BC = x,
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It seems you missed to upload graphs given with the question. So i will draw the graph myself then explain the answer.
f(x)=x^3+8 is the given function which is created from parent function f(x)=x^3
Graph of x^3 is like graph of any odd degree function which starts from bottom and goes to top because of positive leading coefficient.
then there is +8 which means parent function is shifted 8 units upward.
You may directly plug some numbers to find points on f(x)=x^3+8 then join those points to get the graph easily.
like we can use numbers x=-2,-1,0,1,2,...
I will show you calculation for x=-2
f(-2)=(-2)^3+8=(-2)(-2)(-2)+8=-8+8=0
Hence first point is (-2,0)
Similarly you can find more points then use them to find the final graph as shown in attached picture.
