Question

Consider two parabolas: One has equation 1 ( 4)( 4) 2 y x x =−+ . The other has the same xintercepts, but goes through the point (2,−12) How far apart are the vertices of the two parabolas

Answer 1

Following are the responses to the given question:

Explanation:

`\to y=((1)/(2))(x-4)(x+4)\\\\\to y=((1)/(2)) (x^2-16)\\\\\to y=((1)/(2))(x-0)^2-8\\\\vertex \to (0,-8)`

The general x-intercept parabola equation
`y=k(x-4)(x+4)`

Parabola crosses the dot (2,-12)

`\to k(2-4)(2+4)=-12\\\\\to k(-2)(6)=-12\\\\\to -12k=-12\\\\\to k=(-12)/(-12)\\\\\to k=1`

The parabolic equation which crosses the position
`(2,-12)` is
`y=(x-4)(x+4)`

`\to y=(x-4)(x+4)\\\\\to y=x^2-16\\\\\to y=(x-0)^2-16\\\\vertex \to (0,-16)`

The distance among the vertices of the two parabolas:

`= √((0 - 0)^2+(-8-(-16))^2)\\\\ = √(0+(-8+16))^2)\\\\ =√(0+(8)^2)\\\\=√((8)^2)\\\\= 8\\\\`