• Feb 29th 2008, 06:47 PM
delicate_tears
http://i77.photobucket.com/albums/j5...enthesetwo.jpg
Between e, and f how do I know whether to use y= a(x-h)^2+k form, or ax^2 +bx +c form? And how do I work it out?

http://i77.photobucket.com/albums/j5...ars/quest3.jpg
And how do I work out this one step by step?
• Feb 29th 2008, 09:12 PM
earboth
Quote:

Originally Posted by delicate_tears
http://i77.photobucket.com/albums/j5...enthesetwo.jpg
Between e, and f how do I know whether to use y= a(x-h)^2+k form, or ax^2 +bx +c form? And how do I work it out?

http://i77.photobucket.com/albums/j5...ars/quest3.jpg
And how do I work out this one step by step?

to #3:

All points on the x-axis have the y-coordinate zero. Thus you know 2 points of the parabola:

P(-1, 4), Q(6, 0) . Plug in these coordinates into the given equation. You'll get a system of linear equations. Solve for a and b:

$\left|\begin{array}{l}4 = a - b\\0 = 36a+6b\end{array}\right.$ ...... I've got $(a, b) = \left(\frac47, -\frac{24}{7}\right)$
• Feb 29th 2008, 09:18 PM
earboth
Quote:

Originally Posted by delicate_tears
http://i77.photobucket.com/albums/j5...enthesetwo.jpg
Between e, and f how do I know whether to use y= a(x-h)^2+k form, or ax^2 +bx +c form? And how do I work it out?

...

If you know the coordinates of the vertex the equation $y= a(x-h)^2+k$ would be the best.

In both cases you know the coordinates of the vertex and the coordinates of one additional point. I'm going to do f so you can try to do e:

You know: V(2, 2) and P(0, 6). Plug in these coordinates:

$6 = a (0 - 2)^2 + 2~\implies~ 6 = 4a+2~\implies~a = 1$ . Thus your equation becomes:

$y = (x-2)^2+2$
• Mar 19th 2008, 04:09 PM
rbell35
Quadratic Rules - h & k values
Does this mean that h & k will always be the x & y co-ordinates of the turning point, if the quadratic is written in the form y = a(x-h) squared + k??? Please clarify..Thanks, Rob