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SectionThe Vertex of a Parabola

The graph of a quadratic function \(f(x) = ax^2 + bx + c\) is called a parabola. Some parabolas are shown in Figure299.

two parabolas with labeled features
Figure299

All these parabolas share certain features. The graph has either a highest point (if the parabola opens downward, as in Figure299a) or a lowest point (if the parabola opens upward, as in Figure299b). This high or low point is called the vertex of the graph.

The parabola is symmetric about a vertical line, called the axis of symmetry, that runs through the vertex. The \(y\)-intercept is the point where the parabola intersects the \(y\)-axis. The graph of a quadratic function always has exactly one \(y\)-intercept.

However, the graph may cross the \(x\)-axis at one point, at two points, or not at all. Points where the parabola intersects the \(x\)-axis are called the \(x\)-intercepts. If there are two \(x\)-intercepts, they are equidistant from the axis of symmetry.

The values of the constants \(a\text{,}\) \(b\text{,}\) and \(c\) determine the location and orientation of the parabola. We will begin by considering each of these constants separately.

Vertex Form

When a quadratic function is written in the form

\begin{equation*} y=a(x-h)^2+k \end{equation*}

we say that the quadratic is written in vertex form. Once written in this form, the vertex of the parabola is the point \((h,k)\) and the axis of symmetry is the line \(x=h \text{.}\)

Caution300

Notice in the definition given above, the vertex of the parabola is \((\alert{h},k)\) and yet in the formula we have \(-h \) showing up

\begin{equation*} y=a(x\alert{-h})^2+k. \end{equation*}

If, for example we have a parabola written as

\begin{equation*} y=2(x+4)^2-3 \end{equation*}

then the vertex is not \((4,-3)\text{,}\) but rather it is \((-4,-3)\) as we would have to rewrite it as

\begin{equation*} 2(x+4)^2-3=2(x-(-4))^2-3 \end{equation*}

in order to have the parabola written in vertex form.

SubsectionFinding the Vertex

The process we will use to put a quadratic into vertex form is through a process called completing the square. Given a quadratic function in the form

\begin{equation*} f(x)=x^2+bx, \end{equation*}

to complete the square we simply

  1. divide \(b \) by \(2\) and square the outcome, to get

    \begin{equation*} \left(\dfrac{b}{2}\right)^2 \end{equation*}

  2. then "add zero" to the function by adding and subtracting this number

    \begin{equation*} f(x)=x^2+bx=\left(x^2+bx+\left(\dfrac{b}{2}\right)^2\right)-\left(\dfrac{b}{2}\right)^2 \end{equation*}

  3. we have then completed the square in that we may write \begin{align*} f(x)=x^2+bx \amp =\alert{\left(x^2+bx+\left(\dfrac{b}{2}\right)^2\right)}-\left(\dfrac{b}{2}\right)^2\\ \amp = \alert{\left(x+\dfrac{b}{2}\right)^2}-\left(\dfrac{b}{2}\right)^2 \end{align*}

The best way to learn this is through examples, so let's work through one together.

Example301

Put the quadratic \(y=2x^2 - 6x - 5\) into vertex form.

Solution

  1. Start by factoring out the coefficient of the term \(x^2\text{,}\) but do not factor it out of the constant term:

    \begin{equation*} 2x^2 - 6x - 5=2\cdot(\alert{x^2 - 3x}) -5. \end{equation*}

    Now, rewrite the expression within the parenthesis with the constant on the right side.

    \begin{equation*} 2\cdot(\alert{x^2 - 3x} ~\underline{\hspace{2.727272727272727em}} )-5 \end{equation*}

  2. Complete the square within the parenthesis, leaving the coefficient of \(2\) and the constant of \(-5\) unchanged:

    \begin{equation*} p = \frac{1}{2}(-3) = \frac{-3}{2} ~~~\text{ and } ~~~ p^2 = \left(\frac{-3}{2}\right)^2 = \frac{9}{4} \end{equation*}

    Add and subtract \(\frac{9}{4}\text{:}\)

    \begin{equation*} 2\cdot (x^2 - 3x \alert{+\frac{9}{4}} -\alert{+\frac{9}{4}}) -5. \end{equation*}

  3. Rewrite as the square of a binomial:

    \begin{equation*} 2\cdot\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)-5 \end{equation*}

  4. Finally, we simplify to get the function in vertex form by distributing the \(2\) and combining like terms:

    \begin{equation*} 2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}-5 \end{equation*}
    \begin{equation*} 2\left(x-\frac{3}{2}\right)^2-\frac{19}{2} \end{equation*}

SubsectionThe Graph of \(y = ax^2 + bx + c\)

Consider the function

\begin{equation*} y = 2x^2 + 8x + 6. \end{equation*}

If we put this quadratic into vertex form we get the equation

\begin{equation*} y=2\cdot (x+2)^2-2. \end{equation*}

From the vertex form we read off the vertex to be \((-2,-2)\) with axis of symmetry along the line \(x=-2\text{.}\)

We find the \(x\)-intercepts of the graph by setting \(y\) equal to zero. For this, it is easier to go back to the function as written in standard form: \begin{align*} 0 \amp= 2x^2 + 8x + 6\amp\amp\text{Factor the right side.}\\ \amp= 2(x + 1)(x + 3)\amp\amp\text{Set each factor equal to zero.}\\ x + 1 \amp=0 ~~~\text{ or } ~~~x + 3 = 0\\ x\amp= -1 \hphantom{blankbl} x = -3 \end{align*} The \(x\)-intercepts are the points \((-1, 0)\) and \((-3, 0)\text{.}\)

The \(y\)-intercept of the graph is found by setting \(x\) equal to zero:

\begin{equation*} y = 2(\alert{0})2 + 8(\alert{0}) + 6 = 6 \end{equation*}

You can see that the \(y\)-coordinate of the \(y\)-intercept, \((0,6)\text{,}\) is just the constant term of the quadratic equation. The completed graph is shown in Figure302.

Figure302
Example303

Find the vertex of the graph of \(y = -2x^2 + x + 1\text{.}\)

Solution

For this equation, \(a = -2\text{,}\) \(b = 1\text{,}\) and \(c = 1\text{.}\) The \(x\)-coordinate of the vertex is given by

\begin{equation*} x_v =\frac{-b}{2a}=\frac{-1}{2(-2)}=\frac{1}{4}. \end{equation*}

To find the \(y\)-coordinate of the vertex, we substitute \(x = \alert{\frac{1}{4}}\) into the equation. We can do this by hand to find \begin{align*} y_v\amp= -2\left(\alert{\frac{1}{4}}\right)^2+ \alert{\frac{1}{4}}+1\\ \amp= -2\left(\frac{1}{16}\right)+ \frac{4}{16}+\frac{16}{16}=\frac{18}{16} =\frac{9}{8}. \end{align*} So the coordinates of the vertex are \(\left(\dfrac{1}{4},\dfrac{9}{8}\right)\text{.}\) Alternatively, we can use the calculator to evaluate \(-2x^2 + x + 1\) for \(x = 0.25\text{.}\) The calculator returns the \(y\)-value \(1.125\text{.}\) Thus, the vertex is the point \((0.25, 1.125)\text{,}\) which is the decimal equivalent of \(\left(\dfrac{1}{4},\dfrac{9}{8}\right)\text{.}\)

SubsectionSketching a Parabola

Once we have located the vertex of the parabola, the \(x\)-intercepts, and the \(y\)-intercept, we can sketch a reasonably accurate graph. Recall that the graph should be symmetric about a vertical line through the vertex. We summarize the procedure as follows.

To Graph the Quadratic Function \(y = ax^2 + bx + c\text{:}\)

  1. Determine whether the parabola opens upward (if \(a \gt 0\)) or downward (if \(a \lt 0\)).

  2. Locate the vertex of the parabola.

    1. The \(x\)-coordinate of the vertex is \(x_v =\dfrac{-b}{2a}\text{.}\)

    2. Find the \(y\)-coordinate of the vertex by substituting \(x_v\) into the equation of the parabola.

  3. Locate the \(x\)-intercepts (if any) by setting \(y = 0\) and solving for \(x\text{.}\)

  4. Locate the \(y\)-intercept by evaluating \(y\) for \(x = 0\text{.}\)

  5. Locate the point symmetric to the \(y\)-intercept across the axis of symmetry.

Example304

Find the vertex of the graph of \(y = 3x^2 - 6x + 4\text{.}\) Decide whether the vertex is a maximum point or a minimum point of the graph.

Solution

For this equation, \(a=3\text{,}\) \(b=-6\text{,}\) and \(c=4\text{.}\) The \(x\)-coordinate of the vertex is given by

\begin{equation*} x_v=\frac{-b}{2a}=\frac{-(-6)}{2(3)}=1 \end{equation*}

To find the \(y\)-coordinate of the vertex, we substitute \(x=\alert{1}\) into the equation. We find

\begin{equation*} y_v=3(1)^2-6(1)+4=3-6+4=1 \end{equation*}

So the coordinates of the vertex are \((1,1)\text{.}\) Since \(a>0\text{,}\) the parabola opens upward and therefore the vertex is a minimum.

Example305

Sketch a graph of the equation \(y = x^2 + 3x + 1\text{,}\) showing the significant points.

Solution

We follow the steps outlined above.

  1. Because \(a = 1 \gt 0\text{,}\) we know that the parabola opens upward.

  2. Compute the coordinates of the vertex:

    \begin{equation*} \begin{aligned} x_v\amp=\frac{-b}{2a}=\frac{-3}{2(1)}= -1.5\\ y_v\amp= (-1.5)2 + 3(-1.5) + 1 = -1.25 \end{aligned} \end{equation*}

    The vertex is the point \((-1.5,-1.25)\text{.}\)

  3. Set \(y\) equal to zero to find the \(x\)-intercepts.

    \begin{equation*} \begin{aligned} 0 \amp= x^2 + 3x + 1\amp\amp\text{Use the quadratic formula.}\\ x \amp=\frac{-3\pm\sqrt{3^2-4(1)(1)}}{2(1)}\\ \amp=\frac{-3\pm\sqrt{5}}{2} \end{aligned} \end{equation*}

    Rounding to the nearest tenth, we find that the \(x\)-intercepts are approximately \((-2.6, 0)\) and \((-0.4, 0)\text{.}\)

  4. Substitute \(x = 0\) to find the \(y\)-intercept, \((0, 1)\text{.}\)

  5. The axis of symmetry is the vertical line \(x = -1.5\text{,}\) so the \(y\)-intercept lies \(1.5\) units to the right of the axis of symmetry.

    There must be another point on the parabola with the same \(y\)-coordinate as the \(y\)-intercept but \(1.5\) units to the left of the axis of symmetry. The coordinates of this point are \((-3, 1)\text{.}\)

    Finally, plot the \(x\)-intercepts, the vertex, and the \(y\)-intercept and its symmetric point, and draw a parabola through them. The finished graph is shown in Figure306.

    parabola
    Figure306

SubsectionExercises