Question

which of the following is the graph of $f(x) = 3|x - 4| + 1$?

Explanation:

Step1: Identify the vertex of the absolute - value function

The general form of an absolute - value function is \(y = a|x - h|+k\), where \((h,k)\) is the vertex of the V - shaped graph. For the function \(f(x)=3|x - 4|+1\), comparing with the general form, we have \(h = 4\) and \(k = 1\). So the vertex of the graph should be at the point \((4,1)\).

Step2: Analyze the direction and the slope

Since \(a = 3>0\), the graph opens upwards (the V - shape opens upwards). The slope of the right - hand side of the V (for \(x\geq h\)) is \(a = 3\), and the slope of the left - hand side (for \(x < h\)) is \(-a=- 3\).

Step3: Analyze the graphs based on vertex and direction

• For the first graph: The vertex is at a negative \(x\) - value, so it does not match the vertex \((4,1)\).
• For the second graph: The vertex is at \((4, - 1)\) (approximate) and the graph opens downwards (since the arms of the V are pointing down), but our function has \(a = 3>0\) (opens upwards) and vertex \((4,1)\), so it does not match.
• For the third graph: The vertex is at a negative \(x\) - value, so it does not match the vertex \((4,1)\).
• For the fourth graph (the bottom - most graph): The vertex is at \(x = 4\) (we can check the position of the vertex on the \(x\) - axis) and \(y=1\) (approximate from the grid), and the graph opens upwards, which matches the properties of \(f(x)=3|x - 4|+1\) (vertex \((4,1)\) and \(a = 3>0\) (opens upwards)).

Answer:

The bottom - most graph (the fourth graph in the given set of graphs)