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

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

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Answer 1 · 2018-05-08T09:18:41

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

$2 {(0)}^{4} - 5 {(0)}^{2} = - 3 y + 12$ $2(0)^4-5(0)^2= -3y + 12$

now solve for y:

$0 = - 3 y + 12$ $0 = -3y + 12$

add $3 y$ $3y$ on both sides

$3 y = 12$ $3y = 12$

divide both sides by $3$ $3$

$y = 4$ $y = 4$

$y-intercept point (0, 4)$ $color(red)("y-intercept point" (0, 4))$

for x-intercept replace $y$ $y$ by $0$ $0$

So

$2 x^{4} - 5 x^{2} = - 3 (0) + 12$ $2x^4-5x^2 =-3(0)+12$

solve for x:

$2 x^{4} - 5 x^{2} = 12$ $2x^4 - 5x^2 = 12$

$2 x^{4} - 5 x^{2} - 12 = 0$ $2x^4 - 5x^2 - 12= 0$

$let x^{2} = x$ $"let" x^2 = x$

$2 x^{2} - 5 x - 12 = 0$ $2x^2 - 5x - 12= 0$

factor

$2 x^{2} - 8 x + 3 x - 12 = 0$ $2x^2 - 8x +3x - 12= 0$

--here I find two numbers their product is -24(because of $2 \cdot - 12$ $2*-12$ )and their sum is -5
and replace them in -5x place--

common factor

$2 x (x - 4) + 3 (x - 4) = 0$ $2x(x-4)+3(x-4)=0$

$(2 x + 3) (x - 4) = 0$ $(2x+3)(x-4)=0$

$2 x + 3 = 0$ $2x+3=0$ and $x - 4 = 0$ $x-4=0$

$x = - \frac{3}{2}$ $x = -3/2$ and $x = 4$ $x=4$

now remember we've changed $x^{2}$ $x^2$ by $x$ $x$
so:

$x^{2} = - \frac{3}{2}$ $x^2=-3/2$ and $x^{2} = 4$ $x^2=4$

$x^{2} = - \frac{3}{2}$ $x^2=-3/2$ is rejected because of exponential can not equal to negative

$x^{2} = 4$ $x^2 = 4$ sequare both sides $x = \pm \sqrt{4}$ $x = +-sqrt4$
$x = 2$ $x = 2$ or $x = - 2$ $x = -2$

$x-intercept points (2, 0), (- 2, 0)$ $color(red)("x-intercept points" (2,0) , (-2,0)$

Answer 2 · 2018-05-08T09:20:13

$x-intercepts = \pm 2, y-intercept = 4$ $"x-intercepts "=+-2," y-intercept "=4$

Explanation:

$to find the intercepts, that is where the graph crosses$ $"to find the intercepts, that is where the graph crosses"$
$the x and y axes$ $"the x and y axes"$

$∙ let x = 0, in the equation for y-intercept$ $• " let x = 0, in the equation for y-intercept"$

$∙ let y = 0, in the equation for x-intercepts$ $• " let y = 0, in the equation for x-intercepts"$

$x = 0 \Rightarrow - 3 y = - 12 \Rightarrow y = 4 \leftarrow y-intercept$ $x=0rArr-3y=-12rArry=4larrcolor(red)"y-intercept"$

$y = 0 \Rightarrow 2 x^{4} - 5 x^{2} - 12 = 0$ $y=0rArr2x^4-5x^2-12=0$

$use the substitution u = x^{2}$ $"use the substitution "u=x^2$

$\Rightarrow 2 u^{2} - 5 u - 12 = 0$ $rArr2u^2-5u-12=0$

$using the a-c method to factor$ $"using the a-c method to factor"$

$the factors of the product 2 \times - 12 = - 24$ $"the factors of the product "2xx-12=-24$

$which sum to - 5 are - 8 and + 3$ $"which sum to - 5 are - 8 and + 3"$

$split the middle term using these factors$ $"split the middle term using these factors"$

$\Rightarrow 2 u^{2} - 8 u + 3 u - 12 = 0 \leftarrow factor by grouping$ $rArr2u^2-8u+3u-12=0larrcolor(blue)"factor by grouping"$

$2 u (u - 4) + 3 (u - 4) = 0$ $2u(u-4)+3(u-4)=0$

$\Rightarrow (u - 4) (2 u + 3) = 0$ $rArr(u-4)(2u+3)=0$

$change u back into terms in x$ $"change u back into terms in x"$

$\Rightarrow (x^{2} - 4) (2 x^{2} + 3) = 0$ $rArr(x^2-4)(2x^2+3)=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$2 x^{2} + 3 = 0 \Rightarrow x^{2} = - \frac{3}{2} \leftarrow no real solutions$ $2x^2+3=0rArrx^2=-3/2larrcolor(blue)"no real solutions"$

$x^{2} - 4 = 0 \Rightarrow x^{2} = 4$ $x^2-4=0rArrx^2=4$

$\Rightarrow x = - 2 or x = + 2 \leftarrow x-intercepts$ $rArrx=-2" or "x=+2larrcolor(red)"x-intercepts"$
graph{-2/3x^4+5/3x^2+4 [-10, 10, -5, 5]}

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

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