Lesson 17

Here are graphs representing the functions \(f\) and \(g\), given by \(f(x) = x(x+6)\) and \(g(x) = x(x+6)+4\).

1. Which graph represents each function? Explain how you know.
2. Where does the graph of \(f\) meet the \(x\)-axis? Explain how you know.

1. How would you change the equation \(y=x^2\) so that the vertex of the graph of the new equation is located at the following coordinates and the graph opens as described?
   1. \((0,11)\), opens up
   2. \((7,11)\), opens up
   3. \((7,\text-3)\), opens down
2. Use graphing technology to verify your predictions. Adjust your equations if necessary.

3. Kiran graphed the equation \(y=x^2+1\) and noticed that the vertex is at \((0,1)\). He changed the equation to \(y=(x-3)^2+1\) and saw that the graph shifted 3 units to the right and the vertex is now at \((3,1)\).

   Next, he graphed the equation \(y= x^2 +2x +1\), observed that the vertex is at \((\text-1,0)\). Kiran thought, “If I change the squared term \(x^2\) to \((x-5)^2\), the graph will move 5 units to the right and the vertex will be at \((4,0)\).”

   Do you agree with Kiran? Explain or show your reasoning.

Mai is learning to create computer animation by programming. In one part of her animation, she uses a quadratic function to show the path of the main character, an animated peanut, jumping over a wall.

Mai uses the equation \(y = \text-0.1(x-h)^2+k\) to represent the path of the jump. \(y\) represents the height of the peanut as a function of the horizontal distance it travels \(x\). On the screen, the base of the wall is located at \((22, 0)\), with the top of the wall at \((22,4.5)\).

1. What are the values of \(k\) and \(h\) in this equation?
2. This applet is programmed to use Mai’s equation. Experiment with the applet to find an equation to get the peanut over the wall, and keep it on the screen. Be prepared to explain your reasoning.

Do you see 2 “eyes” and a smiling “mouth” on the graph? The 3 arcs on the graph all represent quadratic functions that were initially defined by \(y=x^2\), but whose equations were later modified.

 

Notice that adding 12 to \(x^2\) raises the graph by 12 units, so the vertex of that graph is at \((0,12)\). Replacing \(x^2\) with \((x+3)^2\) shifts the graph 3 units to the left, so the vertex is now at \((\text-3,0)\).

We can also shift a graph both horizontally and vertically.