Question

Which is the equation of the parabola?

coordinate plane with a parabola facing up with vertex at 4 comma 2, the point 4 comma 6 and a horizontal line going through 0 comma negative 2


y = one sixteenth(x − 4)2 + 2

y = one sixteenth(x + 4)2 − 2

y = −one sixteenth(x − 2)2 + 4

y = −one sixteenth(x + 2)2 − 4

Answer 1

Answer: The correct answer is the first answer, a. y = one sixteenth(x − 4)2 + 2

Answer 2

Answer: y = one sixteenth(x − 4)^2 + 2

Explanation:

If the parabola is written as:

y = a*x^2 + b*x + c

then if the graph opens up, then a must be positive, so we can discard the third and fourth options, we remain with:

y = 1/6*(x - 4)^2 + 2 = 1/6x^2 - (8/6)*x + (16/6 + 2)

y = 1/6*(x + 4)^2 - 2 = 1/6x^2 + (8/6)*x + (16/6 - 2)

the vertex (4, 2)

then

x = -b/2a = 4.

this means that a and b must be of different sign, then the only correct option can be:

y = 1/6*(x - 4)^2 + 2 = 1/6x^2 - (8/6)*x + (16/6 + 2)

where:

x-vertex = (8/6)/(2/6) = 4 as we wanted.

when we evaluate this function in x = 4 we get

y = 1/6*( 4 - 4)^2 + 2 = 2.

So the correct option must be: y = one sixteenth(x − 4)2 + 2