quadratic function

[heatherbang]

k i odnt know where to put this but:

Write an equation in the form y = bx^2 + bx + c for the quadratic function whose graph passes through points (8,0) (0,8) and (-2,0)

 

[tkhunny]

k i odnt know where to put this but:

Write an equation in the form y = bx^2 + bx + c for the quadratic function whose graph passes through points (8,0) (0,8) and (-2,0)

First, you should notice that it will be only by luck (or problem design) that you will find a solution. With three points, but only two parameters, it could be a problem. Are you SURE you didn't mean y = ax^2 + bx + c.

In any case, just plug and chug.

(8,0) suggests 0 = b(8)^2 + b(8) + c
(0,8) suggests 8 = b(0)^2 + b(0) + c

 

[mowlick]

k i odnt know where to put this but:

Write an equation in the form y = bx^2 + bx + c for the quadratic function whose graph passes through points (8,0) (0,8) and (-2,0)

I think tkhunny is right and that the question should read:

y = ax^2 + bx + c

In that case you have a simultaneous equation:

At point (8,0), x = 8 and y = 0

therefore

y = ax^2 + bx + c =

0 = y = a8^2 + 8b + c = 64a + 8b + c

At point (0,8), x = 0 and y = 8

therefore

y = ax^2 + bx + c =

8 = a(0)^2 + (0)b + c = c

At point (-2,0), x = -2 and y = 0

therefore

y = ax^2 + bx + c =

0 = a(-2)^2 + (-2)b + c = 4a -2b + c

So we now have three equations

(i) 0 = 64a + 8b + c

(ii) 8 = c

(iii) 0 = 4a - 2b + c


from (ii) we know that 8 = c

therefore, (i) and (iii) become:

(i) 0 = 64a + 8b + 8

(iii) 0 = 4a - 2b + 8

If we multiply (iii) by 4 we will end up with

0 = 16a - 8b + 32

Call the above equation (iv)

therefore

(i) 0 = 64a + 8b + 8

(iv) 0 = 16a - 8b + 8

(i) + (iv) =

0 = 80a + 16

therefore

-16 = 64a

therefore

-16/64 = a = -1/4

using (i),

0 = 64a + 8b + 8 =

0 = 64 (-1/4) + 8b + 8 =

0 = -16 + 8b + 8 =

0 = 8b - 8

therefore

8 = 8b

therefore

1 = b

therefore

y = ax^2 + bx + c =

y = (-1/4) x^2 + x + 8