Question

Fit a quadratic function to these three points: (-2,8), (0, -4), and (4, 68) Respuesta

Answer 1

Answer:

f(x) = 4x^2 + 2x - 4.

Step-by-step explanation:

Let the quadratic function be y = f(x) = ax^2 + bx + c.

For the point (-2, 8)  ( x = -2 when y = 8) we have:

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

4a - 2b + c = 8      For (0, -4) we have:

0 + 0 + c = -4   so c = -4.    For (4, 68) we have:

16a + 4b + c = 68  

So we have 2 systems of equations in a and b ( plugging in c = -4):

4a - 2b - 4 = 8

16a + 4b - 4 = 68

4a - 2b = 12

16a + 4b = 72    Multiplying 4a - 2b = 12 by 2 we get:

8a - 4b = 24  

Adding the last 2 equations:

24a = 96

a = 4

Now plugging a = 4 and c = -4 in the first equation:

4(4) - 2b - 4 = 8

-2b = 8 - 16 + 4 = -4

b = 2.