Find the equation of the parabola that passes through the points.

Question

Could you help me with this?

 

Find the equation of the parabola
y = ax2 + bx + c
that passes through the points:
(0, 0), (4, −4), (8, 0)

Answer 1

We can create a system of equations using these points to solve for a, b, and c.

 

Since the point (0, 0) is the y-intercept and the point (8, 0) is the x-intercept, the equations we get are

 

c = 0              eq1     ---->  y-intercept

 

64a + 8b + c = 0           eq2   ----> x-intercept

 

 

Using the other point, we get the equation

 

16a + 4b + c = -4          eq3      ---->   point (4, -4)

 

 

Since c=0, we reduce the equations to

 

64a + 8b = 0                 eq2

16a + 4b = -4                eq3

 

Multiply eq3 by 2.  Keep eq2.

 

64a + 8b = 0                  eq2

32a + 8b = -8                 eq3

 

 

Subtract eq3 from eq2.

 

32a = 8

 

   a = 1/4

 

 

Substitute this value of a into eq2 to solve for b.

 

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

 

16 + 8b = 0

 

        8b = -16

 

         b = -2

 

 

So your equation will be

 

y = (1/4)x² - 2x