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

##### 2 Answers

To find the y-intercepts you substitute 0 as x value

So

now solve for y:

add

divide both sides by

for x-intercept replace

So

solve for x:

factor

**--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

now remember we've changed

so:

#### 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]}