so, the turning point is at
plug in and you will have 2 equations..
for the third equation:
note:
a point is on the curve if and only if ..
The graph of the function y = ax² + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.
I know this is only a basic question, but I really would appreciate your help.
Thanks
This aproach should give you the same answer but you will only have to solve a symoltaneous equation in two variables.
You want the equation to pass through (0,5) so substitute x=0, y=5 into the equation to get y = 5 = c
You want the equation to pass through (-3,2) so substitute x=-3, y=2 into the equation to get
Then by knowing that at the turning point the differential of y with respect to x is zero you can write:
You have two equations in two variables which can be solved by symoltaneous equations.
y=ax^2+bx+c
since it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.
Now at turning point x=-b/2a
(if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5
First use the turning point form y = a (x - h)^2 + k. Obviously h = 0, k = 5. Substitute (-3, 2) to get a.
Now expand the turning point form to get the standard form.
Edit: Don't double post! It wastes peoples time! http://www.mathhelpforum.com/math-he...nt-thanks.html