• anonymous
Describe the differences between the graph of y = –3(x + 7)2 – 10 and the standard position graph of y = x2. Explain how the a value of -3 and the h & k values affect the graph.
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
So, to begin, recall:\[y=a(x-h)^{2}+k\]Where (h,k) is the location of the vertex. With yours:\[y=-3(x+7)^{2}-10\]Lets start by adjusting the h values slightly so we can see what's really going on there.\[y=-3(x-(-7))^{2}-10\]Now, we can officially identify the vertex as (-7,-10). So, the value of -7 affects the graph by causing the vertex (and the whole graph for that matter) to shift to the left 7 units. The value of -10 causes the graph to shift down 10 units. The 'a' value has 2 things that it's doing. First, because it is negative, it causes the graph to open downwards. Secondly, the 3 causes the graph to grow faster so it appears to stretch the parabola with respect to y. This makes the parabola look tall and skinny compared with the function y=x^2

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