Question

What is the equation of a parabola that has intercepts of x=-2 and x=6 and passes through point (8,6)?

Answer 1

Step-by-step explanation:

general form of a parabola :

y = ax² + bx + c

we have 3 points of the parabola to calculate the 3 variables :

the 2 intercepts are actually

(-2, 0)

(6, 0)

and then

(8, 6)

so, we get the 3 equations

0 = a(-2)² + b×-2 + c = 4a - 2b + c

0 = a×6² + b×6 + c = 36a + 6b + c

6 = a×8² + b×8 + c = 64a + 8b + c

from the 1st equation we get

c = -4a + 2b

using that in the 2nd equation we get

0 = 36a + 6b - 4a + 2b = 32a + 8b = 4a + b

b = -4a

therefore,

c = -4a + 2b = -4a + 2×-4a = -4a - 8a = -12a

using all that in the 3rd equation we get

6 = 64a + 8×-4a + -12a = 64a - 32a - 12a = 20a

a = 6/20 = 3/10 = 0.3

b = -4a = -4×3/10 = -12/10 = -1.2

c = -12a = -12×3/10 = -36/10 = -3.6

so, we have

y = 3/10 x² - 12/10 x - 36/10 = 3/10 × (x² - 4x - 12)