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

2. $y = ax^2 + bx + c = x^2 + \frac{b}{a}x + \frac{c}{a}$

$= x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{4ac- b^2}{4a^2} = \left({x +\frac{b}{2a}}\right)^2 + \frac{4ac- b^2}{4a^2}$

so, the turning point is at $\left({-\frac{b}{2a}, \frac{4ac-b^2}{4a^2}}\right)$

plug in and you will have 2 equations..

for the third equation:
note:
a point $(x_0,y_0)$ is on the curve $y = ax^2 + bx + c$ if and only if $y_0 = ax_0^2 + bx_0 + c$..

3. Originally Posted by tim_mannire
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 $2 = 9*a-3b+5$

Then by knowing that at the turning point the differential of y with respect to x is zero you can write:

$y'=0=2*a*x + b$

You have two equations in two variables which can be solved by symoltaneous equations.

4. ## Here 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

5. Originally Posted by tim_mannire
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 would really appreciate any help on this problem, thanks.
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

6. Originally Posted by tim_mannire
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 would really appreciate any help on this problem, thanks.
Use the following formula:

$y = a(x - p)^2 + q$

$y = a(x + 3)^2 + 2$

Passes through the point $(0;5)$

$5 = a(0 + 3)^2 + 2$

$3 = 9a$

$a = \frac{1}{3}$

$y = \frac{1}{3} (x+3)^2 + 2$

EDIT: Oops Mr F beat me to it... by 2 mins

7. Originally Posted by nikhil
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

Thank you very much. very well answered. much appreciated!