Quadratic functions and equations?

Question

Some types of worms have a remarkable capacity to live without moisture. The table shows the number of worms y surviving after x days in one study.

X (days) 0 20 40 80 120 160
Y (worms) 50 48 45 36 20 3

(A) Find a quadratic function in the form
f(x) = a(x-h)^2 + K that models these data.

(B) Solve the quadratic equation f(x) = 0. Do both solutions have real meaning? Explain

Answer 1

OK, so this is gonna be messy.

Since y = f(x) = a(x-h)² + K, if we plug in values for x and y, we still have 3 variables left: a, h, and K.

If we have at least 3 points, however, we can find 3 equations in 3 unknowns, which will get us values for a, h, and K.

So, choose the points (0, 50), (40, 45) and (120, 20) for example. (You could choose any 3; I chose these because they're all multiples of 5, so maybe they'll be easier to simplify.)

Then:

(0, 50):
50 = a(0 - h)² + K = ah² + K

(40, 45):
45 = a(40 - h)² + K = 1600a - 80ah + ah² + K

(120, 20):
20 = a(120 - h)² + K = 14400a - 240ah + ah² + K

So now we have:
50 = ah² + K
45 = 1600a - 80ah + ah² + K
20 = 14400a - 240ah + ah² + K

Messy! But, notice that the 2nd and 3rd equations end with "ah² + K"...and the first equation tells us that ah² + K = 50. So it gets easier:

45 = 1600a - 80ah + 50
-5 = 1600a - 80ah
and
20 = 14400a - 240ah + 50
-30 = 14400a - 240ah

That's a little better at least. Now you'll just have to solve one of the equations for a variable and plug it in the other one. I choose to solve the first equation for h:

-5 - 1600a = -80ah
h = 1/(16a) + 20

Plug this in for h in the second eq:

-30 = 14400a - 240a(1/(16a) + 20)
-30 = 14400a - 15 - 4800a
-15 = 9600a
a = -1/640

Then h = 1/(16(-1/640)) + 20 = -40 + 20 = -20

And finally K = 50 - ah² = 50 - (-1/640)(-20)²
= 50 + 5/8 = 50 5/8

So the final function would be:

f(x) = -1/640(x + 20)² + 50 5/8

Now...I checked this equation with the 3 points I chose, and it fits those 3 points exactly; however, the other points don't fit exactly. So maybe you were supposed to find a "best fit" quadratic, I dunno. Kind of tired of this one.

In any case, taking that equation, setting it equal to 0 and solving, you'd find that x = 160 should correspond to f(x) = 0, and some other root where x is negative. That solution doesn't have real meaning, since the experiment starts at x = 0.

Answer 2

This parabola ought to have easily one of two equations; a million) even if it truly is of the type y = a(x - h)^2 + ok the position the vertex is given as (a million,0) then y = a(x - a million)^2. when you consider that we are given also that (0,2) is on the curve then 2 = a(0 - a million)^2 = a, and so the equation to that end is y = 2(x - a million)^2. 2) even if it truly is of the type x = a(y - ok)^2 + h with vertex (h,ok) = (a million,0) then we've x = a(y - 0)^2 + a million = a(y)^2 + a million. with aspect (0,2) we've 0 = a*2^2 + a million ----> 4a = -a million ---> a = -a million/4, giving the equation x = (-a million/4)y^2 + a million.