# find values for x

• Nov 14th 2009, 07:27 PM
scrible
find values for x
Hi, I have a problem that trying to solve all evning and it is in two parts (a) If $a$ is a positive constant, find the set of valus of $x$for which $a(x^2 +2x-8)$ is negative. Find the value of a if this function has a minimum value of -27.
(b) Find two quadratic funcions of $x$ which are zero at $x=1$, which take the value of 10 when $x=0$and which have a maximum value of 18. Sketch the graps of these two functions. Can ou help me with this problem?
• Nov 14th 2009, 08:14 PM
topsquark
Quote:

Originally Posted by scrible
If $a$ is a positive constant, find the set of valus of $x$for which $a(x^2 +2x-8)$ is negative. Find the value of a if this function has a minimum value of -27.

If a is positive then we need to find where $x^2 + 2x - 8 < 0$. Start by finding out where the trinomial is zero. I get these to be at x = -4 and x = 2. It is simple to see that the negative range for this is -4 < x < 2.

Can you get the rest from here? (Hint: Consider where the parabola has its absolute minimum.

-Dan
• Nov 14th 2009, 08:38 PM
topsquark
Quote:

Originally Posted by scrible
(b) Find two quadratic funcions of $x$ which are zero at $x=1$, which take the value of 10 when $x=0$and which have a maximum value of 18. Sketch the graps of these two functions.

You are speaking of a quadratic. The general form is
$y = ax^2 + bx + c$

For x = 1 we have a zero, so y = 0.
$0 = a + b + c$

When x = 0 we have y = 10, so
$10 = c$

$y = ax^2 + bx + 10$
where $a + b = -10$
$x_{max/min} = -\frac{b}{2a}$
You can plug that into the $y = ax^2 + (-a - 10)x + 10$ equation to get
$18 = ax^2 + (-a - 10)x + 10$