quadratic curve

**poniekid** · May 17th 2008, 05:37 AM

ImageShack - Hosting :: mathdo0.jpg

sorry for the scribbling.
could someone please show me how to do this.
thanks

**galactus** · May 17th 2008, 06:13 AM

Here's a graph of two such parabolas.

Since they cross the y-axis at 0, then c=0.

**earboth** · May 17th 2008, 06:29 AM

Originally Posted by poniekid

ImageShack - Hosting :: mathdo0.jpg

sorry for the scribbling.
could someone please show me how to do this.
thanks

Use the given equation to calculate the missing coefficients:

Parabola 1:
$y=-2x^2+bx+c$

P(-2,0) and Q(0, 0)

Plug in the coefficients into the equation:

$Q: 0=0+0+c~\implies~ c=0$

$P: 0=-2\cdot (-2)^2+b\cdot(-2)+0~\implies~ b= -4$ and therefore the equation of parabola1 becomes:

$p_1:y=-2x^2-4x$

Do just the same to calculate the equation of parabola2. You should come out with

$p_2:y=-2x^2+4x$

The domain is [-2, 0] or [0, 2] respectively.

By the way: You can get one function whose graph is the required "M":

$y = -2x^2+4\cdot |x|~,~x \in [-2, 2]$

The height of the letter is the y-coordinate of the vertex of one parabola. I don't know which method you prefer to calculate the vertex but you should get: $V_1(-1, 2)$

To answer #5 I've sketched a family of curves:

$y_a=-ax^2+2a \cdot |x|$ for a running from 2 to 6 in steps of 0.5. Now make your choice!

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