# quadratic function problem

• Oct 14th 2010, 05:03 AM
mastermin346
Diagram shows the curve of a quadratic function $f(x) = ax^2 + bx + c$, where $a$, $b$ and $c$ are constant.

a)find the value of $a$, $b$ and $c$

b)Determine the range value of $x$ if $f(x) \geq 1$
• Oct 14th 2010, 05:12 AM
mr fantastic
Quote:

Originally Posted by mastermin346
Diagram shows the curve of a quadratic function $f(x) = ax^2 + bx + c$, where $a$, $b$ and $c$ are constant.

a)find the value of $a$, $b$ and $c$

b)Determine the range value of $x$ if $f(x) \geq 1$

From the point (0, -2): -2 = c.

Update the model: y = ax^2 + bx - 2.

From the point (-1, -3): -3 = a - b - 2.

From the turning point (-1, -3): -1 = -b/(2a).

Solve simultaneously for a and b.

(b) cannot be done until you have correctly answered (a).
• Oct 14th 2010, 05:15 AM
greatersanta616
The point at which the curve crosses the 'y' axis is the value of c. This being -2.
by using the complete the square form, we can work out the rest of the equation.
To do this, we use the vertex, or lowest point of the graph.
y=(x+1)^2-3
this gives us x^2+2x+1-3 or x^2+2x-2
Thus. The values are ....

I cannot recall an answer to part b. But I hope this has helped :)
• Oct 14th 2010, 05:52 AM
mastermin346
hi!

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

$=ax^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b^2}{4}\right) +c$

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

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

$x + \frac{b}{2} = 0$

$-1 + \frac{b}{2} = 0$

$b= 2$

minimum value = $-3$

$-\frac{b^2}{4} + c = -3$

$c=-2$

the value of $a$ is

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

at point $(-1,-3)$

$-3 = a + 2(-1) - 2$

$-3 = a - 4$

$a = 1$

is correct sir?
• Oct 14th 2010, 08:07 PM
Educated
Seems correct to me.

You can always double check by putting in x=-1 and seeing if the y co-ordinate is correct.
• Oct 14th 2010, 08:57 PM
mastermin346
how about the range?
• Oct 14th 2010, 09:05 PM
Educated
$x^2 + 2x - 2 \geq 1$

$x^2 + 2x - 3 \geq 0$

Can you carry on?
• Oct 14th 2010, 09:18 PM
mastermin346
so,the range is $f(x) \geq 1$ and $f(x) \leq -3$?
• Oct 14th 2010, 09:20 PM
Educated
Yep.