Lesson 17

Changing the Vertex

  • Let’s write new quadratic equations in vertex form to produce certain graphs.

17.1: Graphs of Two Functions

Two graphs representing functions in x y plane, origin O.

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.

17.2: Shifting the Graph

  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.

17.3: A Peanut Jumping over a Wall

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 \((22,4.5)\).

The dashed curve in the picture shows the graph of one equation Mai tried, where the peanut fails to make it over the wall.

Graph of the path of a peanut’s jump.
  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.

17.4: Smiley Face

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.

  1. Write equations to represent each curve in the smiley face.
  2. What domain is used for each function to create this graph?
Left eye, Opens down, Vertex at -2 comma 50. Right eye, opens down, vertex = 2 comma 50. Mouth, opens up, vertex = 0 comma 10.

 

Summary

The graphs of \(y = x^2\), \(y = x^2 + 12\) and \(y = (x+3)^2\) all have the same shape but their locations are different. The graph that represents \(y=x^2\) has its vertex at \((0,0)\).

Three parabolas in x y plane, origin O.

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.

The graph that represents \(y = (x+3)^2 +12\) will look like that for \(y = x^2\) but it will be shifted 12 units up and 3 units to the left. Its vertex is at \((\text-3,12)\).

Two parabolas in x y plane, origin O.

The graph representing the equation \(y = \text{-}(x+3)^2 + 12\) has the same vertex at \((\text-3,12)\), but because the squared term \((x+3)^2\) is multiplied by a negative number, the graph is flipped over horizontally, so that it opens downward.

Two parabolas in x y plane, origin O.

Glossary Entries

  • vertex form (of a quadratic expression)

    The vertex form of a quadratic expression in \(x\) is \(a(x-h)^2 + k\), where \(a\), \(h\), and \(k\) are constants, and \(a\) is not 0.