Question

1. saul is analyzing the graph f(x) below and claims he can transform the function so that it includes the ordered pair (1, 6). write an example of a function that saul could have used.

Explanation:

Step1: Identify the original function

The graph is a parabola opening downward with vertex at \((4, 6)\) and roots at \(x = 1\) (wait, no, looking at the graph, the roots seem to be around \(x = 1\)? Wait, no, the graph crosses the x - axis? Wait, no, the graph touches the x - axis? Wait, no, the graph has a vertex at \((4, 6)\) and passes through \((1, - 2)\)? Wait, no, let's re - examine. The graph is a parabola, let's assume the original function is \(f(x)=- (x - 4)^2+6\). Let's check \(x = 1\): \(f(1)=-(1 - 4)^2+6=-9 + 6=-3\). We want to transform it so that when \(x = 1\), \(y = 6\). One way is to use a vertical shift or a horizontal shift or a stretch/compression. Let's use a horizontal shift. Suppose we shift the graph to the left by \(h\) units. The new function is \(y=-(x-(4 - h))^2+6\). We want when \(x = 1\), \(y = 6\). So \(6=-(1-(4 - h))^2+6\). Then \(-(1-(4 - h))^2=0\), so \(1-(4 - h)=0\), \(1-4 + h=0\), \(h = 3\). So the new function is \(y=-(x-(4 - 3))^2+6=-(x - 1)^2+6\). Let's check \(x = 1\): \(y=-(1 - 1)^2+6=6\), which works. Another way: vertical shift. Let the transformed function be \(f(x)=- (x - 4)^2+6 + k\). We want \(f(1)=6\), so \(6=- (1 - 4)^2+6 + k\), \(6=-9 + 6 + k\), \(k = 9\). So \(f(x)=-(x - 4)^2+15\). But a simpler transformation: let's use a horizontal shift. Or we can use a different approach. Let's consider the function \(y = 6\) when \(x = 1\). A constant function? No, but the original is a parabola. Wait, the problem says "transform the function", so we can use a translation. Let's take the original function (assuming vertex form). The original vertex is \((4,6)\). If we shift the graph 3 units to the left, the new vertex is \((1,6)\), so the function is \(y=-(x - 1)^2+6\). When \(x = 1\), \(y=6\), which is the ordered pair \((1,6)\).

Step2: Verify the transformed function

For the function \(y=-(x - 1)^2+6\), when \(x = 1\), \(y=-(1 - 1)^2+6=6\), so the ordered pair \((1,6)\) is on the graph. Also, the shape is a parabola opening downward, which is a transformation of the original parabola (shifted 3 units to the left).

Answer:

\(y =-(x - 1)^2+6\) (or other valid transformed functions like \(y = 6\) (constant function, but since the original is a parabola, a quadratic function is more appropriate. Another example: \(y=-(x - 4)^2+15\) also works when \(x = 1\): \(y=-(1 - 4)^2+15=-9 + 15 = 6\))