Question

How do I summarize the x and y intercepts of 2x^4-12x^2 ?

Answer 1

for y intercept is easy: imagine some graph that goes round and round. The graph will cross the y-axis when x=0, or when you substitute into the function y(x)=2x^4-12x^2 with x=0, you get that the y-intercept is at the origin (0,0).

Answer 2

by the same logic, the graph intercepts the x-axis when y=0, that is you have to solve the equation 0=2x^4-12x^2 (just substitute y=0 to get it), so when you divide by 2 you get:

0=x^4-6x^2
0=x^2(x^2-6)
then use the formula a^2-b^2=(a-b)*(a+b)

x^2*(x-sqrt(6))*(x+sqrt(6))=0

then you solve the three equations x^2=0 =>x=0;
x-sqrt(6)=0 => x=sqrt(6)
and by the same logic for the last x-intercept you get x=-sqrt(6),

where sqrt(something) stands for square root of something.