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f(x)={[-(1)/(2)x-4," if "x < -1],[(x-2)^(2)," if "-1 <= x < 3],[|x-5|-3," if "x >= 3]:}

f(x)={−12x−4 if x<−1(x−2)2 if −1≤x<3∣x−5∣−3 if x≥3 f(x)=\left\{\begin{array}{cl}-\frac{1}{2} x-4 & \text { if } x<-1 \\ (x-2)^{2} & \text { if }-1 \leq x<3 \\ |x-5|-3 & \text { if } x \geq 3\end{array}\right.

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Full solution

Q. f(x)={−12x−4 if x<−1(x−2)2 if −1≤x<3∣x−5∣−3 if x≥3 f(x)=\left\{\begin{array}{cl}-\frac{1}{2} x-4 & \text { if } x<-1 \\ (x-2)^{2} & \text { if }-1 \leq x<3 \\ |x-5|-3 & \text { if } x \geq 3\end{array}\right.
  1. Identify Piecewise Function: Identify the piecewise function and its components. f(x)f(x) is defined differently for different intervals of xx.
  2. Find Quadratic Vertex: Find the vertex of the quadratic piece.\newlineFor the interval −1≤x<3-1 \leq x < 3, f(x)=(x−2)2f(x) = (x-2)^2.\newlineThe vertex of f(x)=(x−2)2f(x) = (x-2)^2 is (2,0)(2, 0).
  3. Find Absolute Value Vertex: Find the vertex of the absolute value piece.\newlineFor the interval x≥3x \geq 3, f(x)=∣x−5∣−3f(x) = |x-5|-3.\newlineThe vertex of f(x)=∣x−5∣−3f(x) = |x-5|-3 is (5,−3)(5, -3).
  4. Linear Piece: The linear piece does not have a vertex.\newlineFor the interval x<−1x < -1, f(x)=−(12)x−4f(x) = -(\frac{1}{2})x - 4, which is a line and does not have a vertex.

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