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

y = one sixteenth(x − 4)^2 + 2

Step-by-step 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

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)