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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 Figure289.

All these parabolas share certain features. The graph has either a highest point (if the parabola opens downward, as in Figure289a) or a lowest point (if the parabola opens upward, as in Figure289b). 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 $x=h \text{.}$

###### Caution290

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

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.

###### Example291

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 Figure292. ###### Example293

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.

###### Example294

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.

###### Example295

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 Figure296.