The curve C has the equation
$y = x^2 + ax + b$
where a and b are constants.
Given that the minimum point of C has coordinates (−2, 5), find the values of a and b.
I don't get how to do this question, cause I have missed classes when we were doing this and even after looking at examples in my book still can't understand
can anyone explain ?

thanks

2. Originally Posted by Tweety
I don't get how to do this question, cause I have missed classes when we were doing this and even after looking at examples in my book still can't understand
can anyone explain ?

thanks
The minimum occurs when:

$\frac{d}{dx}\left[x^2+ax+b\right]=2x+a=0$

that is when $x=-a/2$. At $x=-a/2$: $y=a^2/4 - a^2/2 +b=-a^2/4+b$.

But you are told that the point $(-a/2, -a^2/4+b)$ is $(-2,5)$ , so solve for $a$ abd $b$.

CB

3. Without using calculus, completing the square:

$x^2+ ax+ b= x^2+ ax+ \frac{a^2}{4}- \frac{a^2}{4}+ b$[/tex]= (x- a/2)^2+ b- \frac{a^2}{4}[/tex]
Since a square is always positive, that has its minimum value, $b- \frac{a^2}{4}$ when x= a/2.

Your are told that x= a/2= -2 and $y= b- \frac{a^2}{4}= 5$, the same equations Captain Black got.

4. Originally Posted by HallsofIvy
Without using calculus, completing the square:

$x^2+ ax+ b= x^2+ ax+ \frac{a^2}{4}- \frac{a^2}{4}+ b$[/tex]= (x- a/2)^2+ b- \frac{a^2}{4}[/tex]
Since a square is always positive, that has its minimum value, $b- \frac{a^2}{4}$ when x= a/2.

Your are told that x= a/2= -2 and $y= b- \frac{a^2}{4}= 5$, the same equations Captain Black got.
The question is in the calculus sub-forum.

CB