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If f(x-3)=x^2-4x+5, what's f(1-x)​

1 Answer

6 votes

Answer:


\large\boxed{f(1-x)=x^2-4x+5}

Explanation:


f(x-3)=x^2-4x+5\\\\f((x-3)+3)\to\text{put x + 3 to the equation of the function:}\\\\f(x-3+3)=(x+3)^2-4(x+3)+5\\\text{use the distributive property and}\ (a+b)^2=a^2+2ab+b^2\\f(x)=x^2+2(x)(3)+3^2+(-4)(x)+(-4)(3)+5\\f(x)=x^2+6x+9-4x-12+5\qquad\text{combine like terms}\\f(x)=x^2+(6x-4x)+(9-12+5)\\f(x)=x^2+2x+2\\\\f(1-x)\to\text{put 1 - x to the equation of the funtion f(x):}\\\\f(1-x)=(1-x)^2+2(1-x)+2\\\text{use the distributive property and}\ (a-b)^2=a^2-2ab+b^2


f(1-x)=1^2-2(1)(x)+x^2+(2)(1)+(2)(-x)+2\\f(1-x)=1-2x+x^2+2-2x+2\qquad\text{combine like terms}\\f(1-x)=x^2+(-2x-2x)+(1+2+2)\\f(1-x)=x^2-4x+5

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