Question

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

Answer 1

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)#

Answer 2

`"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]}