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The function, fx) = x^2 - 4x + 3, has y-values that increase when x<2. TrueFalse

User Wei Liu
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Let's begin by listing out the information given to us:


\begin{gathered} f(x)=x^2-4x+3 \\ f(x)=y \\ \Rightarrow y=x^2-4x+3 \\ y=x^2-4x+3 \end{gathered}

We will proceed to choose values for x (values of x lesser than 2); x = 1, 0, -1


\begin{gathered} y=x^2-4x+3 \\ x=1 \\ y=1^2-4(1)+3=1-4+3=4-4=0 \\ y=0 \\ (x,y)=(1,0) \\ \\ x=0 \\ y=0^2-4(0)+3=0-0+3=3 \\ y=3 \\ (x,y)=(0,3) \\ \\ x=-1 \\ y=(-1^2)-4(-1)+3=1+4+3=8 \\ y=8 \\ (x,y)=(-1,8) \end{gathered}

From the calculation, we see a trend that the y-values increase as the x-value decreases. Hence, it is true

User Benqus
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