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Study Guides > College Algebra CoRequisite Course

Characteristics of Parabolas

Learning Outcomes

  • Identify the vertex, axis of symmetry, yy-intercept, and minimum or maximum value of a parabola from it's graph.
  • Identify a quadratic function written in general and vertex form.
  • Given a quadratic function in general form, find the vertex.
  • Define the domain and range of a quadratic function by identifying the vertex as a maximum or minimum.
The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are. The yy-intercept is the point at which the parabola crosses the yy-axis. The xx-intercepts are the points at which the parabola crosses the xx-axis. If they exist, the xx-intercepts represent the zeros, or roots, of the quadratic function, the values of xx at which y=0y=0.

tip for success

The places where a function's graph crosses the horizontal axis are the places where the function value equals zero. You've seen that these values are called horizontal intercepts, x-intercepts, and zeros so far. They can also be referred to as the roots of a function.

Example: Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown below. Graph of a parabola with a vertex at (3, 1) and a y-intercept at (0, 7).

Answer: The vertex is the turning point of the graph. We can see that the vertex is at (3,1)(3,1). The axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is <em>x</em>=3<em>x</em>=3. This parabola does not cross the xx-axis, so it has no zeros. It crosses the yy-axis at (0, 7) so this is the yy-intercept.

Try It

Use an online graphing calculator to help you solve the question below. [ohm_question hide_question_numbers=1]155642[/ohm_question]

Equations of Quadratic Functions

The general form of a quadratic function presents the function in the form

f(x)=ax2+bx+cf\left(x\right)=a{x}^{2}+bx+c

where aa, bb, and cc are real numbers and a≠0a\ne 0. If a>0a>0, the parabola opens upward. If a<0a<0, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. The axis of symmetry is defined by x=−b2ax=-\dfrac{b}{2a}. If we use the quadratic formula, x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}, to solve ax2+bx+c=0a{x}^{2}+bx+c=0 for the xx-intercepts, or zeros, we find the value of xx halfway between them is always x=−b2ax=-\dfrac{b}{2a}, the equation for the axis of symmetry. The figure below shows the graph of the quadratic function written in general form as y=x2+4x+3y={x}^{2}+4x+3. In this form, a=1, b=4a=1,\text{ }b=4, and c=3c=3. Because a>0a>0, the parabola opens upward. The axis of symmetry is x=−42(1)=−2x=-\dfrac{4}{2\left(1\right)}=-2. This also makes sense because we can see from the graph that the vertical line x=−2x=-2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, (−2,−1)\left(-2,-1\right). The xx-intercepts, those points where the parabola crosses the xx-axis, occur at (−3,0)\left(-3,0\right) and (−1,0)\left(-1,0\right). Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are for the function y=x^2+4x+3. The standard form of a quadratic function presents the function in the form

f(x)=a(x−h)2+kf\left(x\right)=a{\left(x-h\right)}^{2}+k

where (h, k)\left(h,\text{ }k\right) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.

Try It

Using an online graphing calculator, plot the function f(x)=2(x−h)2+kf\left(x\right)=2\left(x-h\right)^2+k. Change the values of hh and kk to examine how changing the location of the vertex (h,k)(h,k) of a parabola also changes the axis of symmetry. Notice that when you move kk independently of hh, you are only moving the vertical location of the vertex. Experiment with values between −10-10 and 1010.   The vertex of a parabola is the location of either the maximum or minimum value of the parabola. If a>0a>0, the parabola opens upward and the parabola has a minimum value of kk at x=hx=h. If a<0a<0, the parabola opens downward, and the parabola has a maximum value of kk at x=hx=h. In this case, the vertex is the location of the minimum value of the function because a=2a=2.

Given a quadratic function in general form, find the vertex of the parabola.

One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, kk, and where it occurs, hh. If we are given the general form of a quadratic function:

f(x)=ax2+bx+cf(x)=ax^2+bx+c

We can define the vertex, (h,k)(h,k), by doing the following:
  • Identify aa, bb, and cc.
  • Find hh, the xx-coordinate of the vertex, by substituting aa and bb into h=−b2ah=-\dfrac{b}{2a}.
  • Find kk, the yy-coordinate of the vertex, by evaluating k=f(h)=f(−b2a)k=f\left(h\right)=f\left(-\dfrac{b}{2a}\right)

Example: Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function f(x)=2x2−6x+7f\left(x\right)=2{x}^{2}-6x+7. Rewrite the quadratic in standard form (vertex form).

Answer: The horizontal coordinate of the vertex will be at

\begin{align}h&=-\dfrac{b}{2a}\ \\[2mm] &=-\dfrac{-6}{2\left(2\right)} \\[2mm]&=\dfrac{6}{4} \\[2mm]&=\dfrac{3}{2} \end{align}

The vertical coordinate of the vertex will be at

\begin{align}k&=f\left(h\right) \\[2mm]&=f\left(\dfrac{3}{2}\right) \\[2mm]&=2{\left(\dfrac{3}{2}\right)}^{2}-6\left(\dfrac{3}{2}\right)+7 \\[2mm]&=\dfrac{5}{2}\end{align}

So the vertex is (32,52)\left(\dfrac{3}{2},\dfrac{5}{2}\right) Rewriting into standard form, the stretch factor will be the same as the aa in the original quadratic.  

f(x)=2(x−32)2+52f\left(x\right)=2{\left(x-\frac{3}{2}\right)}^{2}+\frac{5}{2}

Try It

Given the equation g(x)=13+x2−6xg\left(x\right)=13+{x}^{2}-6x, write the equation in general form and then in standard form.

Answer: g(x)=x2−6x+13g\left(x\right)={x}^{2}-6x+13 in general form; g(x)=(x−3)2+4g\left(x\right)={\left(x - 3\right)}^{2}+4 in standard form

Finding the Domain and Range of a Quadratic Function

Any number can be the input value of a quadratic function. Therefore the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum at the vertex, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all yy-values greater than or equal to the yy-coordinate of the vertex or less than or equal to the yy-coordinate at the turning point, depending on whether the parabola opens up or down.

A General Note: Domain and Range of a Quadratic Function

The domain of any quadratic function is all real numbers. The range of a quadratic function written in general form f(x)=ax2+bx+cf\left(x\right)=a{x}^{2}+bx+c with a positive aa value is f(x)≥f(−b2a)f\left(x\right)\ge f\left(-\frac{b}{2a}\right), or [f(−b2a),∞)\left[f\left(-\frac{b}{2a}\right),\infty \right); the range of a quadratic function written in general form with a negative aa value is f(x)≤f(−b2a)f\left(x\right)\le f\left(-\frac{b}{2a}\right), or (−∞,f(−b2a)]\left(-\infty ,f\left(-\frac{b}{2a}\right)\right]. The range of a quadratic function written in standard form f(x)=a(x−h)2+kf\left(x\right)=a{\left(x-h\right)}^{2}+k with a positive aa value is f(x)≥kf\left(x\right)\ge k; the range of a quadratic function written in standard form with a negative aa value is f(x)≤kf\left(x\right)\le k.

How To: Given a quadratic function, find the domain and range.

  1. The domain of any quadratic function as all real numbers.
  2. Determine whether aa is positive or negative. If aa is positive, the parabola has a minimum. If aa is negative, the parabola has a maximum.
  3. Determine the maximum or minimum value of the parabola, kk.
  4. If the parabola has a minimum, the range is given by f(x)≥kf\left(x\right)\ge k, or [k,∞)\left[k,\infty \right). If the parabola has a maximum, the range is given by f(x)≤kf\left(x\right)\le k, or (−∞,k]\left(-\infty ,k\right].

Example: Finding the Domain and Range of a Quadratic Function

Find the domain and range of f(x)=−5x2+9x−1f\left(x\right)=-5{x}^{2}+9x - 1.

Answer: As with any quadratic function, the domain is all real numbers or (−∞,∞)\left(-\infty,\infty\right). Because aa is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the xx-value of the vertex.

h=−b2a=−92(−5)=910h=-\dfrac{b}{2a}=-\dfrac{9}{2\left(-5\right)}=\dfrac{9}{10}

The maximum value is given by f(h)f\left(h\right).

f(910)=5(910)2+9(910)−1=6120f\left(\dfrac{9}{10}\right)=5{\left(\dfrac{9}{10}\right)}^{2}+9\left(\dfrac{9}{10}\right)-1=\dfrac{61}{20}

The range is f(x)≤6120f\left(x\right)\le \dfrac{61}{20}, or (−∞,6120]\left(-\infty ,\dfrac{61}{20}\right].

Try It

Find the domain and range of f(x)=2(x−47)2+811f\left(x\right)=2{\left(x-\dfrac{4}{7}\right)}^{2}+\dfrac{8}{11}.

Answer: The domain is all real numbers. The range is f(x)≥811f\left(x\right)\ge \dfrac{8}{11}, or [811,∞)\left[\dfrac{8}{11},\infty \right).

[ohm_question]120300[/ohm_question]

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