determine the function f whose graph is given the vertex is (3,-13) and the y intercept is -4 how whould i work this out
I am wondering what degree the funcion is: second degree, fifth degree?
second
Since you are using the word "vertex" I will assume you might be talking about a quadratic function.
with vertex (3,-13) the parabola would be
y = a(x-3)^2 -13
but (0,-4) lies on it, so ....
-4 = a(9) - 13
9a = 9
a = 1
f(x) = (x-3)^2 - 13
thank you so much but i am still not understanding how you got that
in your text or in your notes, you should have the usual form of the equation if you have the vertex
if vertex is (h, k)
then the equation of the quadratic is
y = a(x-h)^2 + k
compare the values of (h,k) with your vertex of (3,-13)
the value of a govern which way the parabola opens and how "steep" or "shallow" it is.