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

2 Answers
May 8, 2018

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)#

May 8, 2018

Answer:

#"x-intercepts "=+-2," y-intercept "=4#

Explanation:

#"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]}