PCC SLC Math Resources

For each of the parabolic equations given below, state the values of \(h\) and \(k\) as well as the coordinates of the vertex of the parabola. Check your answer using the Desmos graph.

1. \(y=(x-4)^2-3\)
2. \(y=(x+2)^2-4\) (Hint: Adding \(2\) to \(x\) is equivalent to subtracting what value from \(x\text{?}\))
3. \(y=(x+4)^2\)
4. \(y=x^2+3\)
5. \(y=-(x-5)^2+2\)
6. \(y=2(x+\frac{3}{2})^2-\frac{5}{2}\)

In the Desmos graph, deactivate "Show vertex" and activate "Show axis of symmetry." For any given parabola whose equation has the form \(y=a(x-h)^2+k\text{,}\) the vertical line \(x=h\) is called the axis of symmetry for the parabola. One way to think about the axis of symmetry is that if you were able to fold your screen across the line, the two halves of the parabola formed by the line would lie atop one another.

For each of the equations given below, state the value of \(h\) and in the Desmos graph slide the value of \(h\) accordingly. Without changing the value of \(h\text{,}\) slide the values of \(a\) and \(k\) back and forth and note that whatever their values, the resultant parabola is symmetric across the line \(x=h\)

1. \(y=a(x-4)^2+k\)
2. \(y=a(x+3)^2+k\)
3. \(y=x^2+k\)

In the Desmos graph, deactivate both "Show vertex" and "Show axis of symmetry" by clicking on the circles to their lefts (if necessary). Activate the equation below "Enter a parabolic equation ..." by clicking on the circle to the left of the equation. Slide the value of \(a\) to \(1\) and leave it there for this entire activity.

The standard form equation stated in the Desmos graph is initially \(y=x^2-6x+4\text{.}\) Note that the vertex of the resultant parabola is located at the point \((3,-5)\text{.}\) This means that in the vertex form of the equation, \(h=3\) and \(k=-5\text{;}\) use the sliders in the Desmos graph to verify this.

Because the graphs of \(y=x^2-6x+4\) and \(y=(x-3)^2-5\) are identical, it must be the case that \(x^2-6x+4=(x-3)^2-5\) for all values of \(x\text{.}\) The process of manipulating the standard form expression into the vertex form of the expression is called "completing the square." The value of \(h\) in the vertex form equivalent of the equation \(y=x^2+bx+c\) is always \(-\frac{b}{2}\)

For each equation of form \(y=x^2+bx+c\) given in Table 2.6.2:

1. Enter the standard form equation into the Desmos graph and make note of the vertex.
2. State the values of \(b\) and \(h\) and confirm that \(h=-\frac{b}{2}\text{.}\)
3. Use the Desmos activity to verify that \(y=(x-h)^2+k\) indeed graphs to the same parabola as did the standard form of the equation.
4. By hand, expand and simplify the right side of the vertex form of the equation and confirm that it does in fact simplify to the right side of the standard form equation.

The process of completing the square involves adding a numeric term to an expression of form \(x^2+bx\) so that the resultant trinomial factors into a perfect square. The number that completes the square is always \((\frac{b}{2})^2\) and the resultant factorization is always \((x+\frac{b}{2})^2\text{.}\)

To complete the square on an equation of form \(y=x^2+bx+c\text{,}\) perform the following steps.

1. Group the squared and linear terms:
   \begin{equation*} y=(x^2+bx)+c \end{equation*}
2. Inside the parentheses, add \((\frac{b}{2})^2\) and balance that addition by subtracting \((\frac{b}{2})^2\) outside the parentheses:
   \begin{equation*} y=(x^2+bx+(\frac{b}{2})^2)+c-(\frac{b}{2})^2 \end{equation*}
3. Factor the trinomial expression in the parentheses and combine the numeric terms outside the parentheses:
   \begin{equation*} y=(x+\frac{b}{2})^2+(c+(\frac{b}{2})^2) \end{equation*}

Let's see a couple of examples with actual numbers.

Use the completing the square process to determine the vertex form of the equation of the parabola whose standard form equation is \(y=x^2+6x+7\text{.}\)

Let's begin by noting that \(b=6\text{,}\) \(\frac{b}{2}=3\) and \((\frac{b}{2})^2=9\text{.}\)

The vertex of the parabola lies at the point \((-3,-2)\text{.}\)

Use the completing the square process to determine the vertex form of the equation of the parabola whose standard form equation is \(y=x^2-10x+32\text{.}\)

Let's begin by noting that \(b=-10\text{,}\) \(\frac{b}{2}=-5\text{,}\) and \((\frac{b}{2})^2=25\text{.}\)

The vertex of the parabola lies at the point \((5,7)\text{.}\)

Time for you to try some yourself. Perform the completing the square process for each of the following equations. Once you have the equation in vertex form, state the vertex of the parabola. Use the Desmos graph to check your result.

1. \(y=x^2-8x+12\)
2. \(y=x^2+4x+1\)
3. \(y=x^2+12x+40\)
4. \(y=x^2-3x-\frac{3}{4}\)

Having a coefficient on the squared term that does not equal \(1\) adds quite a bit of complexity to the completing the square process. Rather than stating the steps using letters, let's illustrate them with specific parabolic equations.

Use the completing the square process to determine the vertex form of the equation of the parabola whose standard form equation is \(y=2x^2+12x+7\text{.}\)

1. Group the squared and linear terms and factor \(a\) from those two terms:
   \begin{align*} y\amp=(2x^2+12x)+7\\ y\amp=2(x^2+6x)+7 \end{align*}
2. Complete the square inside the parentheses and balance the addition outside the parentheses. Since every term inside the parentheses is multiplied by \(a\text{,}\) you must multiply what you add inside the parentheses by \(a\) to determine the number that is subtracted outside the parentheses.
   \begin{equation*} y=2(x^2+6x+9)+7-18 \end{equation*}
   To reiterate, because we multiplied the additional 9 by 2, we need to subtract by 18 to compensate for the addition.
3. Factor the trinomial into its perfect square form and state the vertex of the parabola:
   \begin{equation*} y=2(x+3)^2-11 \end{equation*}
   The vertex of the parabola lies at the point \((-3,-11)\text{.}\)

Use the completing the square process to determine the vertex form of the equation of the parabola whose standard form equation is \(y=-x^2+4x-7\text{.}\)

1. Group the squared and linear terms and factor \(a\) from those two terms:
   \begin{align*} y\amp=(-x^2+4x)77\\ y\amp=-(x^2-4x)+7 \end{align*}
2. Complete the square inside the parentheses and balance the addition outside the parentheses. Since every term inside the parentheses is multiplied by \(a\text{,}\) you must multiply what you add inside the parentheses by \(a\) to determine the number that is subtracted outside the parentheses.
   \begin{equation*} y=-(x^2-4x+4)-7+4 \end{equation*}
   That was a little tricky - at first glance it looks like we added \(4\) twice - not good. But there's a negative sign in front of those parentheses, so while we did indeed add \(4\) inside the parentheses, from the big picture perspective we actually subtracted by \(4\) which needed to be balanced by adding \(4\) outside the parentheses.
3. Factor the trinomial into its perfect square form and state the vertex of the parabola:
   \begin{equation*} y=-(x-2)^2-3 \end{equation*}
   The vertex of the parabola lies at the point \((2,-3)\text{.}\)

Time for you to try some yourself. Perform the completing the square process for each of the following equations. Once you have the equation in vertex form, state the vertex of the parabola. Use the Desmos graph to check your result.

1. \(y=3x^2-12x+12\)
2. \(y=-2x^2-4x+1\)
3. \(y=4x^2-24x+30\)
4. \(y=-x^2+6x-7\)
5. \(y=\frac{1}{2}x+4x+2\)