wasangagac4
2022-10-14
Answered

if $f(x+3)=2{x}^{2}-3x-1$ find $f(x+1)$

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zupa1z

Answered 2022-10-15
Author has **20** answers

You should change 2x - 9 too. Therefore,

$\mathit{f}(\mathit{x}+3)=(\mathit{x}+3)(2\mathit{x}-9)+26$

$\mathit{f}(\mathit{x}+3)=(x+3)(2(x+3)-15)+26$

$\mathit{f}(\mathit{x}+1)=(x+1)(2(x+1)-15)+26=(x+1)(2x-13)+26=2{x}^{2}-11\mathit{x}+13$

$\mathit{f}(\mathit{x}+3)=(\mathit{x}+3)(2\mathit{x}-9)+26$

$\mathit{f}(\mathit{x}+3)=(x+3)(2(x+3)-15)+26$

$\mathit{f}(\mathit{x}+1)=(x+1)(2(x+1)-15)+26=(x+1)(2x-13)+26=2{x}^{2}-11\mathit{x}+13$

Chaim Ferguson

Answered 2022-10-16
Author has **4** answers

You have that $f(x+3)=2{x}^{2}-3x-1$. Write $y=x+3$, then $x=y-3$ and thus

$$f(y)=2(y-3{)}^{2}-3(y-3)-1=2{y}^{2}-15y+26.$$

Hence $f(x+1)=2(x+1{)}^{2}-15(x+1)+26=2{x}^{2}-11x+13$

The key step is to first write x+3 as y and change all of the x's to y−3. In this way you rewrite the equation so that you get f(y) in terms of y. Then you can replace y by whatever you like.

$$f(y)=2(y-3{)}^{2}-3(y-3)-1=2{y}^{2}-15y+26.$$

Hence $f(x+1)=2(x+1{)}^{2}-15(x+1)+26=2{x}^{2}-11x+13$

The key step is to first write x+3 as y and change all of the x's to y−3. In this way you rewrite the equation so that you get f(y) in terms of y. Then you can replace y by whatever you like.

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