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smallz1993 · Aug 9, 2009, 07:37 PM

Help. I do not know how to do this problem at all. Can anyone help me get going?

Find a quatric function f(x)=ax2+bx+c whose graph has a min value at 25 and x intercepts -3 and 2,

Unknown008 · Aug 10, 2009, 10:48 AM

Do you know how to complete the square?

If you do, then it's easier.

For this, you have to make the coefficient of x^2 be 1. So, factor a;

$a(x^2+\frac ba x+\frac ca)$

Then, put x^2 and b/a into brackets which is squared. Write x and for the other term, with coefficient halved and without x, then subtract the square of the coefficient (I know that's confusing, but look below to see what happened);

$a((x+\frac {b}{2a})^2- (\frac {b}{2a})^2+\frac ca)$

Simplify;

$a((x+\frac {b}{2a})^2- \frac {b^2}{4a^2}+\frac ca)$

Now, put back the a.

$a(x+\frac {b}{2a})^2- \frac {b^2}{4a}+ c$

The min is 25, so, from the completed squared form, you should know that the part after the squared brackets give the min and max points.

Therefore,

$- \frac {b^2}{4a}+ c = 25$

Then, it meets the x axis at -3 and 2, that means, when y = 0, x = -3 and 2.

Replace in the completed square form;

$a(-3+\frac {b}{2a})^2- \frac {b^2}{4a}+ c = 0$

$a(2+\frac {b}{2a})^2- \frac {b^2}{4a}+ c = 0$

You have three equations, with three unknowns. I think you can solve it now.

Hope it helped! :)