Find expressions for the quadratic functions whose graphs are shown.

**zweevu** · September 7th 2009, 04:40 PM

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

**Prove It** · September 7th 2009, 04:46 PM

Originally Posted by zweevu

Find expressions for the quadratic functions whose graphs are shown.

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

First, note that the y-intercept is 1.

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

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

Equation 1:
$2 = a(-2)^2 + b(-2) + 1$

$4a - 2b + 1 = 2$

$4a - 2b = 1$ .

Equation 2:
$-2.5 = a(1)^2 + b(1) + 1$

$a + b + 1 = -2.5$

$a + b = -1.5$ .

So your system of equations is...

$4a - 2b = 1$

$a + b = -1.5$ .

Solve for a and b.

**pickslides** · September 7th 2009, 04:51 PM

The second function has the form

$y = ax^2+bx+c$

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

$y = ax^2+bx+1$

Now use points (-2,2)

$y = ax^2+bx+1$

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

and (1,-2.5)

$y = ax^2+bx+1$

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

and solve the sytem

**zweevu** · September 7th 2009, 05:15 PM

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!

**pickslides** · September 9th 2009, 02:23 PM

You have $g(x) = y = ax^2+bx+c$

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

can you find $g(x)$ ?

**OliverQ** · September 10th 2009, 05:43 AM

We assume that:
$g(x) = ax^2 + bx + c$
and the graph of the function got through 3 points, they are:
$(-2, 2), (0, 1), (1, -2.5)$

$g(-2) = 4a - 2b + c = 2;$
$g(0) = c = 1$
$g(1) = a + b +c = -2.5$

So:
$a = -1; b = -2.5; c = 1;$
$g(x) = -x^2 - 2.5x + 1$

**kyleu03** · September 10th 2009, 02:52 PM

can anyone figure out f(x) ???

**pickslides** · September 10th 2009, 03:07 PM

$f(x) = 2(x-3)^2$

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