Find a quadratic model for the set of values
(-2, -20), (0, -4), (4, -20)

Question

Find a quadratic  model for the set of values
(-2, -20), (0, -4), (4, -20)

amistre64 · Answer

we have one root stated, and mirrored ys

amistre64 · Answer

the axis of symmetry, the mirrord ys, has to be the middle of -2 and 4

and when x=0, ah, the y intercept

anonymous · Answer

a0^2+b0+c=-4
means c = -4

anonymous · Answer

You can solve this problem by writing 3 equations. 
The general quadratic equation is
$$ax ^{2}+bx+c=0$$
You have 3 points given, can you make the 3 equations?

amistre64 · Answer

y = a(x-1)^2 + k

    0 = a(-5)^2+k
-20 = a(3)^2+k

3 equations with 2 unknowns is one way to see it

anonymous · Answer

Am not gettin it

amistre64 · Answer

what have you tried?

amistre64 · Answer

bah .... i mixed x,y parts on my first one :)

-4  = a(-1)^2+k
-20 = a(3)^2+k

anonymous · Answer

sorrrrrrrrrrrry $$-2x^2-12x-4$$

anonymous · Answer

Lilmoe is this clear? I think we got you confused.

anonymous · Answer

Sorry yea am confuse

amistre64 · Answer

there are many ways to approach a solution so its hard to define a specific method withou knowing what your own thoughts about it are

amistre64 · Answer

in general we have 3 points ($x_1,~y_1$); ($x_2,~y_2$); ($x_3,~y_3$)

we can setup an equation that can focus on one aspect at a time ... 

we know that anything subtracted from itself is equal to 0,  and that anything times 0 is also 0 so we want to make use of these properties:  ($x_1-x$); ($x_2-x$); ($x_3-x$)

we can construct an equation with coefficients: $c_0,~c_1,~c_2,~...,~c_n$ such that all but one unknown coefficient will zero out and leave us with 1 term to find a solution for like this:

$$y=c_0+c_1(x_1-x)+c_2(x_1-x)(x_2-x)$$
when x=x1,  y=y1 and the c1,c2 terms zero out giving us
$$y_1=c_0$$

knowing c0,  we can let x=x2, y=y2 zeroes out the c3

$$y_2=y_1+c_1(x_1-x_2)$$
$$\frac{y_2-y_1}{x_1-x_2}=c_1$$
c_1 is just the linear slope between the first 2 points ... 

letting x=x3 and y=y3 lets us solve the final coefficient
$$y_3=y_1+\frac{y_2-y_1}{x_1-x_2}(x_1-x_3)+c_2(x_1-x_3)(x_2-x_3)$$
$$y_3-y_1-\frac{y_2-y_1}{x_1-x_2}(x_1-x_3)=c_2(x_1-x_3)(x_2-x_3)$$
$$\frac{y_3-y_1}{(x_1-x_3)(x_2-x_3)}-\frac{y_2-y_1}{(x_1-x_2)(x_2-x_3)}=c_2$$