Question

find the equation of parabola which has axis parallel to y-axis and which passes through the point (0,2), (-1,0), and (1,6)?

Answer 1

Please see the attachment

Answer 2

The equation of the parabola is y = x² + 3x + 2

Step-by-step explanation:

Let the equation of the parabola whose axis is parallel to y -axis is given by

y = ax² + bx + c ......... (1)

Now, this equation satisfies the points (0,2), (-1,0) and (1,6).

So, putting x = 0 and y = 2 in equation (1) we get,

2 = 0 + 0 + c

⇒ c = 2

Again, putting x = - 1 and y = 0, we get, 0 = a - b + 2 {As c = 2}

⇒ a - b = - 2 ........ (2)

Now, putting x = 1 and y = 6 in the equation (1) we get,

6 = a + b + 2

⇒ a + b = 4 .......... (3)

Now, solving equations (2) and (3) we get,

2a = 2

⇒ a = 1

And from equation (3) we get, b = 3

Therefore, the equation of the parabola is

y = x² + 3x + 2  (Answer)