Find expressions for the quadratic functions whose graphs are shown.

http://www.webassign.net/scalcet/1-2-8.gif

I know how to do f(x) but I'm stuck on g(x).

- Sep 7th 2009, 04:40 PMzweevuFind expressions for the quadratic functions whose graphs are shown.
Find expressions for the quadratic functions whose graphs are shown.

http://www.webassign.net/scalcet/1-2-8.gif

I know how to do f(x) but I'm stuck on g(x). - Sep 7th 2009, 04:46 PMProve It
First, note that the y-intercept is 1.

So $\displaystyle g(x) = ax^2 + bx + 1$.

Using the other two points, we will have two equations in two unknowns to solve simultanously.

Equation 1:

$\displaystyle 2 = a(-2)^2 + b(-2) + 1$

$\displaystyle 4a - 2b + 1 = 2$

$\displaystyle 4a - 2b = 1$.

Equation 2:

$\displaystyle -2.5 = a(1)^2 + b(1) + 1$

$\displaystyle a + b + 1 = -2.5$

$\displaystyle a + b = -1.5$.

So your system of equations is...

$\displaystyle 4a - 2b = 1$

$\displaystyle a + b = -1.5$.

Solve for a and b. - Sep 7th 2009, 04:51 PMpickslides
The second function has the form

$\displaystyle y = ax^2+bx+c$

The y-intercept c is 1 as shown in the picture so

$\displaystyle y = ax^2+bx+1$

Now use points (-2,2)

$\displaystyle y = ax^2+bx+1$

$\displaystyle 2 = a(-2)^2+b(-2)+1$ ...(1)

and (1,-2.5)

$\displaystyle y = ax^2+bx+1$

$\displaystyle -2.5 = a(1)^2+b(1)+1$ ...(2)

and solve the sytem - Sep 7th 2009, 05:15 PMzweevu
The question is looking g(x) = ... what would I type in for the answer. I got +/- 1/6 for a and b though.

Edit: NVM You plug them into the original equation. Thanks! - Sep 9th 2009, 02:23 PMpickslides
You have $\displaystyle g(x) = y = ax^2+bx+c$

With $\displaystyle c = 1 , b = \frac{-7}{6}$ and $\displaystyle a = \frac{-1}{3}$

can you find $\displaystyle g(x)$ ? - Sep 10th 2009, 05:43 AMOliverQ
We assume that:

$\displaystyle g(x) = ax^2 + bx + c$

and the graph of the function got through 3 points, they are:

$\displaystyle (-2, 2), (0, 1), (1, -2.5)$

$\displaystyle g(-2) = 4a - 2b + c = 2;$

$\displaystyle g(0) = c = 1$

$\displaystyle g(1) = a + b +c = -2.5$

So:

$\displaystyle a = -1; b = -2.5; c = 1;$

$\displaystyle g(x) = -x^2 - 2.5x + 1$ - Sep 10th 2009, 02:52 PMkyleu03
can anyone figure out f(x) ???

- Sep 10th 2009, 03:07 PMpickslides
$\displaystyle f(x) = 2(x-3)^2$