Parabolic Function II
Notice that represents a quadratic function
and its graph is a parabola with vertex (h,k).
- Let f(x)=-3(x-3)2+4, then
- find its vertex, [Vertex is (3,4) ].
- find its y-intercept, [ y-intercept is
- 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 so
- sketch the graph of f. [Use Maple syntax to plot f.]
- Let Then solve the problems mentioned
above.
- Find the quadratic function f(x) which has the vertex (-2,3) and
the graph of f passes through (2,0). [Consider the function 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. so Therefore,
the function we need is (you can
use Maple to check if your graph looks right).
- Find the quadratic function f(x) which has the vertex (2,-3) and
the graph of f passes through (1,1).
- Find the quadratic function f(x) which has the vertex (1,3) and
the graph of f passes through (-2,-3).