The graph of the functiony = ax˛ + bx + chas a turning point at (-3,2) and passes through the point (0,5). Determine the values ofa,bandc.

I know this is only a basic question, but I really would appreciate your help.

Thanks

Printable View

- June 25th 2008, 12:39 AMtim_mannirePlease help me with this problem, it is urgent, thanks!
The graph of the function

*y = ax˛ + 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

- June 25th 2008, 01:01 AMkalagota

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 .. - June 25th 2008, 01:17 AMKiwi_Dave
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. - June 25th 2008, 01:21 AMnikhilHere it is
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 - June 25th 2008, 01:22 AMmr fantastic
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 - June 25th 2008, 01:24 AMjanvdl
- June 25th 2008, 01:26 AMtim_mannire