# Math Help - Modelling a quadratic equation.

1. ## Modelling a quadratic equation.

A quadratic function with equation:

f(x) = ax^2/2 + bx + c

Passes through the points

(0,2) and (2,-6)

And has a gradient of 1 at x=3.

Solve for the constants a, b and c.

I have c, which is 2, however i have no idea how to find the others.
I understand differentiation and know how to do it etc, i am just not sure about how to apply it to this question/ do the simultaneous equations required.

Just for reference, the answers are :

A = 5/2

B= -13/2

C= 2

2. Originally Posted by HelpMeWithMaths
A quadratic function with equation:

f(x) = ax^2/2 + bx + c

Passes through the points

(0,2) and (2,-6)

And has a gradient of 1 at x=3.

Solve for the constants a, b and c.

I have c, which is 2, however i have no idea how to find the others.
I understand differentiation and know how to do it etc, i am just not sure about how to apply it to this question/ do the simultaneous equations required.

Just for reference, the answers are :

A = 5/2

B= -13/2

C= 2
Hi HelpMeWithMaths,

you got "c" by substituting in x=0 and y=2, as y=f(x).

You need "a" and "b" so you need two clues.

You've been given the clues.
If you now use that value of "c", you can use the second point given,
as it gives

$\frac{a(2)^2}{2}+b(2)+2=-6$

If you differentiate the function, you get the gradient,
hence that is your final clue.

You will have a pair of simultaneous equations to solve for "a" and "b".

3. Saying that the curve passes through the point (0, 2) means that when x= 0, y= 2. Putting x= 0, y= 2 into $y= ax^2/2+ bx+ c$ gives $2= a(0^2)/2+ b(0)+ c= c$. That tells you that c= 2 which you got.

Saying that the curve passes through the point (2, -6) means that when x= 2, y= -6. Putting x= 2, y= -5 into $y= =ax^2+ bx+ c$ gives $-6= a(2^2)/2+ b(2)+ c= 2a+ 2b+ c$. Since you already know that c= 2, $-6= 2a+ 2b+ 2$ which is the same as $2a+ 2b= -8$. You might want to divide through by 2: $a+ b= -4$.

You need another equation to be able to solve for both a and b and you are told that the gradient (derivative) at x= 1 is 3. Although this was posted in the "Precalculus" section, you are clearly expected to know that the derivative of $ax^2$ is 2ax, the derivative bx is b, and the derivative of c is 0. That is, you should know that the derivative (gradient) of $ax^2/2+ bx+ c$ is $2ax/2+ b= ax+ b$. Saying that the derivative is 1 when x= 3 tells you that $a(3)+ b= 3a+ b= 1$.

You now two equations, a+ b= -4 and 3a+ b= 1 to solve for a and b. Probably the simplest way to proceed is to subtract one of those equations from the other.

4. Thank you so much !