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Section 6.2 The 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 Figure 6.14.
described in detail following the image
two parabolas with labeled features
Figure 6.14.
All these parabolas share certain features. The graph has either a highest point (if the parabola opens downward, as in Figure 6.14a) or a lowest point (if the parabola opens upward, as in Figure 6.14b). 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 \(x=h \text{.}\)

Caution 6.15.

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.

Subsection 6.2.1 Finding 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.

Example 6.16.

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} ~\fillinmath{XXXXXX} )-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*}

Subsection 6.2.2 The 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 Figure 6.17.
Figure 6.17.

Example 6.18.

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 \(\left(\dfrac{1}{4},\dfrac{9}{8}\right)\text{.}\)

Subsection 6.2.3 Sketching 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.

Example 6.19.

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.

Example 6.20.

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 Figure 6.21.
    described in detail following the image
    parabola
    Figure 6.21.

Exercises 6.2.4 Exercises

1. Properties of Quadratic Functions.

Suppose we have the function \(f(x) = -x^2 + 5 x\text{.}\)
a) Find the \(x\)-intercepts and the \(y\)-intercept of this function. Enter the larger \(x\)-intercept first, if applicable.
\(x\)-intercepts: \((\), \()\text{,}\) \((\), \()\)
\(y\)-intercept: \((\), \()\)
b)What are the domain and range of \(f(x)\text{?}\)
  • Domain: \((-\infty, \infty)\text{,}\) Range: \((-\infty, 6.25]\)
  • Domain: \((-\infty, \infty)\text{,}\) Range: \((-\infty, 2.5]\)
  • Domain: \((-\infty, 2.5]\text{,}\) Range: \((-\infty, \infty)\)
  • Domain: \([-\infty, \infty]\text{,}\) Range: \([-\infty, 6.25]\)
c) Select the graph of \(f\left(x\right)\text{.}\)
A B C D
Hint.
a) To find the \(x\)-intercepts, we will want to factor the equation. We can factor a \((-x)\) out of both terms as a common factor. This gives \(f(x) = -x(x - 5)\text{.}\) Remember that the \(x\)-intercepts occur when \(y=0\) and the \(y\) intercept occurs when \(x = 0\text{.}\)
b) Remember the domain of a function is the set of all possible input values. Are there any input values that would result in the function being undefined?
c) Use the information above about the \(x\)-intercepts, \(y\)-intercept, and domain to help pick the right graph. You can also consider whether the graph would open up or down.

2. Vertex Form.

For each part, find the vertex and the axis of symmetry. Also, answer if each parabola opens upward or downward.
a) \(f(x) = 0.4 (x-7)^2 +8\)
Vertex: \((\) , \()\)
Axis of symmetry:
Opens:
b) \(g(x) = -(x + 4.5)^2 - 7.5\)
Vertex: \((\) , \()\)
Axis of symmetry:
Opens:
Hint.
Recall that given a parabola of form \(y = a(x-h)^2 +k\text{,}\) the vertex of the parabola is \((h,k)\text{,}\) and the axis of symmetry is \(x = h\text{.}\) If \(a\) is positive, then the parabola opens upward, and if \(a\) is negative, then the parabola opens downward. Using this information, since these functions are already in vertex form, we just need to identify the pieces.

3. Formulas from Graphs.

Find formulas in vertex form for the quadratic functions pictured below.
a) \(f(x) =\)
The point indicated on the graph is \((-4, 4)\text{.}\)
b) \(g(x) =\)
The point indicated on the graph is \((4, -2)\text{.}\)
Hint.
Recall that we write a parabola in vertex form as \(y = a(x-h)^2 +k\text{.}\) The vertex of the parabola is \((h,k)\text{,}\) and the axis of symmetry is \(x = h\text{.}\) \(a\) is the vertical stretch of the parabola. If \(a\) is positive, then the parabola opens upward, and if \(a\) is negative, then the parabola opens downward.
Use an extra point on the graph to help you find \(a\) once you plug in the information about the vertex from the graph.

4. Converting to Vertex Form.

Write each of the following quadratic functions in vertex form. Identify the vertex and axis of symmetry in each case.
a) \(j(x) = x^2 - 10 x+ 28\)
\(j(x) =\)
Vertex: \((\) , \()\)
Axis of symmetry:
b) \(k(x) = 2 x^2 + 10 x + 6.5\)
\(k(x) =\)
Vertex: \((\) , \()\)
Axis of symmetry:
c) \(m(t) = 4 t^2 - 27\)
\(m(t) =\)
Vertex: \((\) , \()\)
Axis of symmetry:
d) \(n(t) = t^2 - 8 t+13\)
\(n(t) =\)
Vertex: \((\) , \()\)
Axis of symmetry:
Hint.
Recall the vertex form of a quadratic function:
\begin{equation*} y=a(x-h)^2+k, \end{equation*}
where the point \((h,k)\) is the vertex and the line \(x=h\) is the axis of symmetry.
One way to rewrite a quadratic function in vertex form from standard form is to complete the square. The following link may be useful: https://mathbooks.unl.edu/PreCalculus//Solving-Quadratic-Equations.html

5. Finding the Vertex.

A ball is thrown into the air. Its height, in feet, \(t\) seconds later is given by
\begin{equation*} h(t)= 112 t-16t^2. \end{equation*}
a) Find the vertex of \(h(t)\text{.}\)
\((\) , \()\)
b) After how many seconds does the ball attain its maximum height?
seconds
c) What is its height at that moment?
feet
Hint.
a) We want to find the vertex of the parabola. To do so, we can complete the square. The following link may be useful: https://mathbooks.unl.edu/PreCalculus//Solving-Quadratic-Equations.html
For parts b) and c), think about how you would find the maximum value of a parabola that opens down.

6. Converting to Standard Form.

Suppose a parabola has its axis of symmetry at \(x=13\text{,}\) has a vertical stretch of 3, opens downward, and has its vertex at the point \((13 ,-14)\text{?}\)
What is a formula for the parabola in standard form \(f(x) = ax^2 + bx + c\text{?}\) Enter the values of \(a, b,\) and \(c\) in the appropriate places.
\(f(x) =\) \(x^2 +\) \(x +\)
Hint.
We want a formula for \(f(x)\) in standard form given the above requirements. To accomplish this, we can first write \(f(x)\) in vertex form:
\begin{equation*} f(x)=a(x-h)^2+k, \end{equation*}
where \((h,k)\) is the vertex and \(a\) is the vertical stretch. Make sure to plug in all the correct information and then multiply everything through and combine like-terms to write in standard form.

7. Application of Quadratic Functions.

Suppose a farmer has 280 feet of fencing, and she wants to enclose a section of her ground that borders a river. The area enclosed will be a rectangle, but only three sides will be enclosed by the fence as the other side will be bounded by the river. This question will guide you through finding the maximum area the farmer can enclose.
a) Draw a picture of the situation on a piece of paper. Let \(w\) and \(l\) denote the width and length, respectively, of the enclosed rectangular field, with the width parallel to the river. What is the area of the enclosed region? (Enter your answer as a formula.)
\(A =\)
b) Assuming the farmer uses all 280 feet of available fencing, write an equation that gives the perimeter of the fence around the enclosed space. (Enter your answer as a formula.)
\(280=\)
c) Solve for \(w\) in your equation in part (b).
\(w =\)
d) Plug your result from part (c) into your formula in part (a). You should now have a quadratic equation that uses only one variable. What is it?
\(A(l)=\)
e) Find the vertex of the parabola in part (d).
Vertex: \((\) , \()\)
Is this vertex a minimum or a maximum?
f) What is the maximum area that the farmer can enclose?
Maximum area:
What values of \(l\) and \(w\) should he choose to attain this maximum?
\(l =\)
\(w =\)
Hint.
a) We can think of this situation as a rectangle with three sides. There are two sides across from each other, and one side across from the river. We label the two sides across from each other with \(l\text{,}\) and we label the side across from the river with \(w\text{.}\) What’s the area of this rectangle in terms of \(l\) and \(w\text{?}\)
b) The perimeter of the fence making up the enclosed area can be found by adding all three sides together. We have two sides of length \(l\) and one side of length \(w\text{,}\) and since we are assuming the farmer is using all 280 feet of fence, these three sides will add to a total of 280. Write an equation relating this information.
c) Solving for \(w\) in the equation in part (b).
d) Plug in the formula from part (c) in for \(w\) in our formula in part (a).
e) To find the vertex, we will have to complete the square in the equation we found in part (d). The following link may be useful: https://mathbooks.unl.edu/PreCalculus//Solving-Quadratic-Equations.html
Would the graph of this function open up or down? This will help you determine if the vertex is a maximum or minimum.
f) Think about the information the vertex tells you in this case.
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