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 byx
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

"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]}