![\quad \huge \quad \quad \boxed{ \tt \:Answer }](https://img.qammunity.org/2023/formulas/mathematics/college/nw7w2s2669vizr2rgidn9a2bju58noyesk.png)
![\qquad\displaystyle \tt \rightarrow \: {y= -1/6(x⁴ +2x³-7x²-8x+ 12 }](https://img.qammunity.org/2023/formulas/mathematics/college/kfn35o28hu967jfpyz2an7jzwaazive5n2.png)
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![\large \tt Solution \: :](https://img.qammunity.org/2023/formulas/mathematics/college/zwgnomcyck8880zsiue912lekjvl3qy9ii.png)
The values of x for which curves cuts/touches the x - axis are the roots of that particular polynomial.
And that polynomial can be depicted in form :
![\qquad \tt \rightarrow \: a(x - h1) (x - h2) (x - h3)........ (x - hn) = 0](https://img.qammunity.org/2023/formulas/mathematics/college/wqwzmepygttfk6dw43vhqae4p33eyuit9y.png)
[ where, h1, h2, h3... hn represents roots of that polynomial, and " a " is the stretch of curve]
And by that, we can sort out the roots of given polynomial that are :
Since there are four roots, the least degree polynomial formed will have bi - quadratic polynomial.
And it will be represented as :
![\qquad \tt \rightarrow \: y=a(x - ( - 3)) (x - ( - 2))(x - 1)(x - 2)](https://img.qammunity.org/2023/formulas/mathematics/college/aq4w5dvzmbg5or2twdm9qxmi1zxd018koy.png)
![\qquad \tt \rightarrow \:y=a (x + 3)(x + 2)(x - 2)(x - 1)](https://img.qammunity.org/2023/formulas/mathematics/college/iy34ejvuur3b3ooiugx2x8wqn9fnui1cs3.png)
And it can be further solved to get ~
![\qquad \tt \rightarrow \: y=a(x + 3)( {x}^(2) - 4)(x - 1)](https://img.qammunity.org/2023/formulas/mathematics/college/5fk3vk6w1u7csp8f97u0g433scjo2gzek6.png)
![\qquad \tt \rightarrow \:y= a( {x}^(2) - 4)( {x}^(2) + 2x - 3)](https://img.qammunity.org/2023/formulas/mathematics/college/sn0gdk1sodnqjg0qi5ozss4s3v0n0s3mgq.png)
![\qquad \tt \rightarrow \: y=a( {x}^(4) + 2 {x}^(3) - 3 {x}^(2) - 4 {x }^(2) - 8x + 12)](https://img.qammunity.org/2023/formulas/mathematics/college/wr8zlkbbc3viakztui3f5pmvx5zttvpay3.png)
![\qquad \tt \rightarrow \: y=a({x}^(4) + 2 {x}^(3) - 7{x }^(2) - 8x + 12)](https://img.qammunity.org/2023/formulas/mathematics/college/wlar8pr3e0194rf8sbnzre4vnrlepru0j2.png)
Now, it's time to evaluate the value of a, for that we can just use a point that satifys the curve ( i.e (0 , -2)
plug in the values :
![\qquad\displaystyle \tt \rightarrow \: {-2= a(0⁴ +2(0)³-7(0)²-8(0) + 12 }](https://img.qammunity.org/2023/formulas/mathematics/college/mzvzmxd2otpu0vgxiih7irulmcey7ml69y.png)
![\qquad\displaystyle \tt \rightarrow \: {-2= a(0 +0-0-0 + 12 }](https://img.qammunity.org/2023/formulas/mathematics/college/8kao0z2zifep9xh04srhczofp9ypbymgls.png)
![\qquad\displaystyle \tt \rightarrow \: {-2= a( 12 )}](https://img.qammunity.org/2023/formulas/mathematics/college/z6nuex3957x0hddv8j2c2webcfmpfcm1p1.png)
![\qquad\displaystyle \tt \rightarrow \: {a= -2 ÷ 12 }](https://img.qammunity.org/2023/formulas/mathematics/college/84ijuk45zyaxfyv91pim6guz1u0kdg4rvx.png)
![\qquad\displaystyle \tt \rightarrow \: {a= -1/6}](https://img.qammunity.org/2023/formulas/mathematics/college/xdlbi7a0wlu2hv5qj9x25f7wzdojjiphcs.png)
Therefore, the required equation is :
![\qquad\displaystyle \tt \rightarrow \: {y= -1/6(x⁴ +2x³-7x²-8x+ 12 }](https://img.qammunity.org/2023/formulas/mathematics/college/kfn35o28hu967jfpyz2an7jzwaazive5n2.png)
Answered by : ❝ AǫᴜᴀWɪᴢ ❞