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.

Question

anonymous · Answer

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