Parabolic Function II

Notice that tex2html_wrap_inline15 represents a quadratic function and its graph is a parabola with vertex (h,k).

  1. Let f(x)=-3(x-3)2+4, then

    1. find its vertex, [Vertex is (3,4) ].
    2. find its y-intercept, [ y-intercept is tex2html_wrap_inline27
    3. find its x-intercep(s), [ set y=0, we get -3(x-3)2+4=0, this is equivalent to 3(x-3)2=4 or tex2html_wrap_inline37 so tex2html_wrap_inline39
    4. sketch the graph of f. [Use Maple syntax to plot f.]
  2. Let tex2html_wrap_inline45 Then solve the problems mentioned above.
  3. Find the quadratic function f(x) which has the vertex (-2,3) and the graph of f passes through (2,0). [Consider the function tex2html_wrap_inline55 since vertex is (-2,3), we can write f(x)=a(x+2)2+3. Now y= f(x) passes through (2,0), so substitute x=2, and y=0, i.e. tex2html_wrap_inline71 so tex2html_wrap_inline73 Therefore, the function we need is tex2html_wrap_inline75 (you can use Maple to check if your graph looks right).
  4. Find the quadratic function f(x) which has the vertex (2,-3) and the graph of f passes through (1,1).
  5. Find the quadratic function f(x) which has the vertex (1,3) and the graph of f passes through (-2,-3).