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

May 8, 2018

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

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

now solve for y:

$0 = - 3 y + 12$

add $3 y$ on both sides

$3 y = 12$

divide both sides by $3$

$y = 4$

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

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

So

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

solve for x:

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

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

$\text{let} {x}^{2} = x$

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

factor

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

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

common factor

$2 x \left(x - 4\right) + 3 \left(x - 4\right) = 0$

$\left(2 x + 3\right) \left(x - 4\right) = 0$

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

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

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

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

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

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

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

May 8, 2018

$\text{x-intercepts "=+-2," y-intercept } = 4$

#### Explanation:

$\text{to find the intercepts, that is where the graph crosses}$
$\text{the x and y axes}$

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

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

$x = 0 \Rightarrow - 3 y = - 12 \Rightarrow y = 4 \leftarrow \textcolor{red}{\text{y-intercept}}$

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

$\text{use the substitution } u = {x}^{2}$

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

$\text{using the a-c method to factor}$

$\text{the factors of the product } 2 \times - 12 = - 24$

$\text{which sum to - 5 are - 8 and + 3}$

$\text{split the middle term using these factors}$

$\Rightarrow 2 {u}^{2} - 8 u + 3 u - 12 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$2 u \left(u - 4\right) + 3 \left(u - 4\right) = 0$

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

$\text{change u back into terms in x}$

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

$\text{equate each factor to zero and solve for x}$

$2 {x}^{2} + 3 = 0 \Rightarrow {x}^{2} = - \frac{3}{2} \leftarrow \textcolor{b l u e}{\text{no real solutions}}$

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

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