Question

HELP HELP!!

Answer 1

Answer:

0

Step-by-step explanation:

$p_1:~~y = x^2+2\\p_2:~~y = 3x^2+2\\ \\ V{p_1} = \Big(-\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) = \Big(-\dfrac{0}{2}, -\dfrac{0^2-4\cdot 2}{4}\Big) = \Big(0,2\Big) \\ \\ Vp_2 = \Big(x_V, -\dfrac{\Delta}{4a}\Big) = \Big(0, -\dfrac{0^2-4\cdot 3 \cdot 2}{4\cdot 3}\Big) = \Big(0,2\Big) \\ \\ \\ \text{The distance is }0,~~\text{Because the vertices are equal.}$