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
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]}