# Thread: Determining the values of a, b and c for quadratic function given some limits.

1. ## Determining the values of a, b and c for quadratic function given some limits.

Hey, having some problems with this problem. I honestly have no idea how to go about solving it, which is why I came here.

Question:

Determine the real values of a, b, and c for the quadratic function f(x) = ax^2 + bx + c, a cannot equal 0, that satisfy the conditions: f(0) = 0, 'limit as x approaches 1' = 5, and 'limit as x approaches -2' = 8.

I know that these points are in the function, the y intercept is 0 (don't know how that helps), and it looks like there are no discontinuities in this function.

Any help is greatly appreciated!

- Steve

2. Originally Posted by Kakariki
Hey, having some problems with this problem. I honestly have no idea how to go about solving it, which is why I came here.

Question:

Determine the real values of a, b, and c for the quadratic function f(x) = ax^2 + bx + c, a cannot equal 0, that satisfy the conditions: f(0) = 0, 'limit as x approaches 1' = 5, and 'limit as x approaches -2' = 8.

I know that these points are in the function, the y intercept is 0 (don't know how that helps), and it looks like there are no discontinuities in this function.

Any help is greatly appreciated!

- Steve
Since quadratic functions are continous we know that

$\lim_{x \to a}f(x)=f(a), \forall x \in\mathbb{R}$

Using this fact we get

$0=a0^2+b0+c \iff c=0$
$5=a(1)^2+b(1)+c$
$8=a(-2)^2+b(-2)+c$

Solving this system of equations gives $a=3, b=2, c=0$

$f(x)=3x^2+2x$

3. Originally Posted by TheEmptySet
Since quadratic functions are continous we know that

$\lim_{x \to a}f(x)=f(a), \forall x \in\mathbb{R}$

Using this fact we get

$0=a0^2+b0+c \iff c=0$
$5=a(1)^2+b(1)+c$
$8=a(-2)^2+b(-2)+c$

Solving this system of equations gives $a=3, b=2, c=0$

$f(x)=3x^2+2x$
That makes sense, but how do you go about solving the 'system of equations'?

4. Originally Posted by TheEmptySet
[snip]
$0=a0^2+b0+c \iff c=0$

$5=a(1)^2+b(1)+c$

$8=a(-2)^2+b(-2)+c$

[snip]
Originally Posted by Kakariki
That makes sense, but how do you go about solving the 'system of equations'?
The first one is simple.

Substitute the value of c into the second and third equations and simplify them. Now solve those two equations simultaneously

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### if the roots of the eq ax sq 2bx c and bxsq.- 2√acx b=0 are simultaneously real then prove that bsq.=ac

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