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

Answer:

Following are the responses to the given question:

Step-by-step explanation:

$\to y=(\frac{1}{2})(x-4)(x+4)\\\\\to y=(\frac{1}{2}) (x^2-16)\\\\\to y=(\frac{1}{2})(x-0)^2-8\\\\vertex \to (0,-8)$

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

Parabola crosses the dot (2,-12)

$\to k(2-4)(2+4)=-12\\\\\to k(-2)(6)=-12\\\\\to -12k=-12\\\\\to k=\frac{-12}{-12}\\\\\to k=1$

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:

$= \sqrt{(0 - 0)^2+(-8-(-16))^2}\\\\ = \sqrt{0+(-8+16))^2}\\\\ =\sqrt{0+(8)^2}\\\\=\sqrt{(8)^2}\\\\= 8$