• Sep 9th 2009, 02:00 PM
kyleu03
Find expressions for the quadratic functions whose graphs are shown.
f(x) =
g(x) =

http://www.webassign.net/scalcet/1-2-8.gif
• Sep 9th 2009, 02:07 PM
Jameson
The normal form we work with discussing parabolas is \$\displaystyle y=ax^2+bx+c\$. You need to find what "a" and "c" are. You have two variables and two points to work with. That's enough to solve for each variable.

Make sense?

EDIT: Big mistake, my apologies. See below post to clarify.
• Sep 9th 2009, 03:38 PM
kyleu03
that works for g(x) but not f(x) becuase there is only one point (the 3,0 doesnt seem to work)
• Sep 9th 2009, 03:52 PM
Jameson
Quote:

Originally Posted by kyleu03
that works for g(x) but not f(x) becuase there is only one point (the 3,0 doesnt seem to work)

You have another point, (4,2). See?
• Sep 9th 2009, 03:56 PM
kyleu03
Quote:

Originally Posted by Jameson
You have another point, (4,2). See?

what forumla would i use? the point-slope?
• Sep 9th 2009, 04:04 PM
Jameson
Quote:

Originally Posted by kyleu03
what forumla would i use? the point-slope?

I made a big mistake, my apologies. The normal form should be \$\displaystyle y=ax^2+bx+c\$. There are three variables to solve for, a b and c. So you need three pairs of (x,y) points to be able to solve for a b and c. G(x) gives you three points but for F(x) you have to use the fact that the parabola is symmetrical to get a third point.

Since you have the point (4,2) you can deduce that there is also the point (2,2) on the graph. Now you plug in your three (x,y) pairs into the equation I wrote above and solve for your a, b and c.
• Sep 9th 2009, 04:51 PM
11rdc11
Quote:

Originally Posted by Jameson
I made a big mistake, my apologies. The normal form should be \$\displaystyle y=ax^2+bx+c\$. There are three variables to solve for, a b and c. So you need three pairs of (x,y) points to be able to solve for a b and c. G(x) gives you three points but for F(x) you have to use the fact that the parabola is symmetrical to get a third point.

Since you have the point (4,2) you can deduce that there is also the point (3,2) on the graph. Now you plug in your three (x,y) pairs into the equation I wrote above and solve for your a, b and c.

Hi Jameson, The point should be (2,2)
• Sep 9th 2009, 05:07 PM
Jameson
Quote:

Originally Posted by 11rdc11
Hi Jameson, The point should be (2,2)

Grrr, yes you're right. I wasn't looking at the graph when I typed that last response. Good catch. Thank you.
• Sep 9th 2009, 05:09 PM
11rdc11
Quote:

Originally Posted by Jameson
Grrr, yes you're right. I wasn't looking at the graph when I typed that last response. Good catch. Thank you.

No problem I do the same all the time lol