What are the x an y intercepts of #2x^4 - 5x^2 = -3y +12#?

Question

Eva Adamson · Accepted Answer

To find the y-intercepts you substitute 0 as x value
So #2(0)^4-5(0)^2= -3y + 12# now solve for y: #0 = -3y + 12# add #3y# on both sides #3y = 12# divide both sides by #3# #y = 4# #color(red)("y-intercept point"
(0, 4))# for x-intercept replace #y# by #0# So #2x^4-5x^2 =-3(0)+12# solve for x: #2x^4 - 5x^2 = 12# #2x^4 - 5x^2 - 12= 0# #"let" x^2 = x# #2x^2 - 5x - 12= 0# factor #2x^2 - 8x +3x - 12= 0# --here I find two numbers their product is -24(because of #2*-12#)and their sum is -5
and replace them in -5x place-- common factor #2x(x-4)+3(x-4)=0# #(2x+3)(x-4)=0# #2x+3=0#  and #x-4=0# #x = -3/2# and #x=4# now remember we've changed #x^2# by#x#
so: #x^2=-3/2# and #x^2=4# #x^2=-3/2# is rejected because of exponential can not equal to negative #x^2 = 4# sequare both sides #x = +-sqrt4#
#x = 2# or #x = -2# #color(red)("x-intercept points"
(2,0) , (-2,0)#

Avery Alexander · Answer

#"x-intercepts "=+-2," y-intercept "=4# #"to find the intercepts, that is where the graph crosses"# #"the x and y axes"# #• " let x = 0, in the equation for y-intercept"# #• " let y = 0, in the equation for x-intercepts"# #x=0rArr-3y=-12rArry=4larrcolor(red)"y-intercept"# #y=0rArr2x^4-5x^2-12=0# #"use the substitution "u=x^2# #rArr2u^2-5u-12=0# #"using the a-c method to factor"# #"the factors of the product "2xx-12=-24# #"which sum to - 5 are - 8 and + 3"# #"split the middle term using these factors"# #rArr2u^2-8u+3u-12=0larrcolor(blue)"factor by grouping"# #2u(u-4)+3(u-4)=0# #rArr(u-4)(2u+3)=0# #"change u back into terms in x"# #rArr(x^2-4)(2x^2+3)=0# #"equate each factor to zero and solve for x"# #2x^2+3=0rArrx^2=-3/2larrcolor(blue)"no real solutions"# #x^2-4=0rArrx^2=4# #rArrx=-2" or "x=+2larrcolor(red)"x-intercepts"# graph{-2/3x^4+5/3x^2+4 [-10, 10, -5, 5]}

Julia Adams · Answer

undefined To find the x-intercepts, set y = 0 and solve for x.
To find the y-intercepts, set x = 0 and solve for y.

1. For x-intercepts:
    2x^4 - 5x^2 = 12
    Factor out x^2:
    x^2(2x^2 - 5) = 12
    x^2 = 12 / (2x^2 - 5)

Setting y = 0:
    2x^2 - 5 = 0
    x^2 = 5/2
    x = ±√(5/2)

So, the x-intercepts are (√(5/2), 0) and (-√(5/2), 0).

2. For y-intercepts:
    2x^4 - 5x^2 = 12 - 3y
    3y = 12 when x = 0
    y = 12 / 3
    y = 4

So, the y-intercept is (0, 4).