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The Parabola

Graph parabolas with vertices at the origin.

The Parabola

Graph vertical parabolas. (IA 11.2.1)
Graph horizontal parabolas. (IA 11.2.2)

A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Previously, we learned to graph vertical parabolas from the general form or the standard form using properties. Those methods will also work here.

Vertical Parabolas
	General form \(y=a{x}^{2}+bx+c\)	Standard form \(y=a{(x-h)}^{2}+k\)
Orientation	\(a>0\) up; \(a<0\) down	\(a>0\) up; \(a<0\) down
Axis of symmetry	\(x=-\frac{b}{2a}\)	\(x=h\)

Example

Try it.

Graph \(y=\text{-}{x}^{2}+6x-8\) .

Solution


Since a is \(-1,\) the parabola opens downward.
To find the axis of symmetry, find \(x=-\frac{b}{2a}.\)


	The axis of symmetry is \(x=3.\)

The vertex is on the line \(x=3.\)
Let \(x=3.\)


	The vertex is \((3,1).\)

The y -intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Simplify.
	The y -intercept is \((0,-8).\)
The point \((0,-8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-8).\)	Point symmetric to the y -intercept is \((6,-8).\)

The x -intercept occurs when \(y=0.\)
Let \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x .
	The x -intercepts are \((4,0),(2,0).\)
Graph the parabola.

Graph vertical parabolas.

Try it.

Graph \(y=2{x}^{2}+4x+5\) .

Try it.

Graph \(y=-{\left(x-3\right)}^{2}+5\) .

Condensed — the full section is in OpenStax College Algebra 2e.

Graphing Parabolas with Vertices at the Origin

In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola. See .

Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points \((x,y)\) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.

In Quadratic Functions, we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See . Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.

The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. The endpoints of the latus rectum lie on the curve. By definition, the distance \(d\) from the focus to any point \(P\) on the parabola is equal to the distance from \(P\) to the directrix.

To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former.

Let \((x,y)\) be a point on the parabola with vertex \((0,0),\) focus \((0,p),\) and directrix \(y= -p\) as shown in . The distance \(d\) from point \((x,y)\) to point \((x,-p)\) on the directrix is the difference of the y-values: \(d=y+p.\) The distance from the focus \((0,p)\) to the point \((x,y)\) is also equal to \(d\) and can be expressed using the distance formula.

\[\begin{array}{l}d=\sqrt{{(x-0)}^{2}+{(y-p)}^{2}} \\ \ =\sqrt{{x}^{2}+{(y-p)}^{2}}\end{array}\]

Set the two expressions for \(d\) equal to each other and solve for \(y\) to derive the equation of the parabola. We do this because the distance from \((x,y)\) to \((0,p)\) equals the distance from \((x,y)\) to \((x, -p).\)

\[\sqrt{{x}^{2}+{(y-p)}^{2}}=y+p\]\[\begin{array}{l}{x}^{2}+{(y-p)}^{2}={(y+p)}^{2} \\ {x}^{2}+{y}^{2}-2py+{p}^{2}={y}^{2}+2py+{p}^{2} \\ {x}^{2}-2py=2py \\ \ {x}^{2}=4py\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Writing Equations of Parabolas in Standard Form

In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.

Example

Try it.

What is the equation for the parabola with focus \((-\frac{1}{2},0)\) and directrix \(x=\frac{1}{2}?\)

Solution

The focus has the form \((p,0),\) so the equation will have the form \({y}^{2}=4px.\)

Multiplying \(4p,\) we have \(4p=4(-\frac{1}{2})=-2.\)
Substituting for \(4p,\) we have \({y}^{2}=4px=-2x.\)

Therefore, the equation for the parabola is \({y}^{2}=-2x.\)

Graphing Parabolas with Vertices Not at the Origin

Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated \(h\) units horizontally and \(k\) units vertically, the vertex will be \((h,k).\) This translation results in the standard form of the equation we saw previously with \(x\) replaced by \((x-h)\) and \(y\) replaced by \((y-k).\)

To graph parabolas with a vertex \((h,k)\) other than the origin, we use the standard form \({(y-k)}^{2}=4p(x-h)\) for parabolas that have an axis of symmetry parallel to the x-axis, and \({(x-h)}^{2}=4p(y-k)\) for parabolas that have an axis of symmetry parallel to the y-axis. These standard forms are given below, along with their general graphs and key features.

Example

Try it.

Graph \({(y-1)}^{2}=-16(x+3).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Solution

The standard form that applies to the given equation is \({(y-k)}^{2}=4p(x-h).\) Thus, the axis of symmetry is parallel to the x-axis. It follows that:

the vertex is \((h,k)=(-3,1)\)
the axis of symmetry is \(y=k=1\)
\(-16=4p,\) so \(p=-4.\) Since \(p<0,\) the parabola opens left.
the coordinates of the focus are \((h+p,k)=(-3+(-4),1)=(-7,1)\)
the equation of the directrix is \(x=h-p=-3-(-4)=1\)
the endpoints of the latus rectum are \((h+p,k\pm 2p)=(-3+(-4),1\pm 2(-4)),\) or \((-7,-7)\) and \((-7,9)\)

Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See .

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Applied Problems Involving Parabolas

As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola’s axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See . This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.

Parabolic mirrors have the ability to focus the sun’s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters.

Example

Try it.

A cross-section of a design for a travel-sized solar fire starter is shown in . The sun’s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.

ⓐ Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.
ⓑ Use the equation found in part ⓐ to find the depth of the fire starter.

Solution

ⓐ The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form \({x}^{2}=4py,\) where \(p>0.\) The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have \(p=1.7.\) \[\begin{array}{lllll}{x}^{2}=4py & \begin{array}{llll} & & & \end{array}\text{Standard form of upward-facing parabola with vertex (0,0)} \\ {x}^{2}=4(1.7)y & \begin{array}{llll} & & & \end{array}\text{Substitute 1}\text{.7 for }p. \\ {x}^{2}=6.8y & \begin{array}{llll} & & & \end{array}\text{Multiply}.\end{array}\]
ⓑ The dish extends \(\frac{4.5}{2}=2.25\) inches on either side of the origin. We can substitute 2.25 for \(x\) in the equation from part (a) to find the depth of the dish. \[\begin{array}{ll}\ {x}^{2}=6.8y & \text{Equation found in part (a)}. \\ {(2.25)}^{2}=6.8y & \text{Substitute 2}\text{.25 for }x. \\ \ y\approx 0.74 & \text{Solve for }y.\end{array}\]
The dish is about 0.74 inches deep.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

Parabola, vertex at origin, axis of symmetry on x-axis	\({y}^{2}=4px\)
Parabola, vertex at origin, axis of symmetry on y-axis	\({x}^{2}=4py\)
Parabola, vertex at \((h,k),\) axis of symmetry on x-axis	\({(y-k)}^{2}=4p(x-h)\)
Parabola, vertex at \((h,k),\) axis of symmetry on y-axis	\({(x-h)}^{2}=4p(y-k)\)

Key Concepts

A parabola is the set of all points \((x,y)\) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.
The standard form of a parabola with vertex \((0,0)\) and the x-axis as its axis of symmetry can be used to graph the parabola. If \(p>0,\) the parabola opens right. If \(p<0,\) the parabola opens left. See .
The standard form of a parabola with vertex \((0,0)\) and the y-axis as its axis of symmetry can be used to graph the parabola. If \(p>0,\) the parabola opens up. If \(p<0,\) the parabola opens down. See .
When given the focus and directrix of a parabola, we can write its equation in standard form. See .
The standard form of a parabola with vertex \((h,k)\) and axis of symmetry parallel to the x-axis can be used to graph the parabola. If \(p>0,\) the parabola opens right. If \(p<0,\) the parabola opens left. See .
The standard form of a parabola with vertex \((h,k)\) and axis of symmetry parallel to the y-axis can be used to graph the parabola. If \(p>0,\) the parabola opens up. If \(p<0,\) the parabola opens down. See .
Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Graph \(y=\text{-}{x}^{2}+6x-8\) .

Révèle la réponse


Since a is \(-1,\) the parabola opens downward.
To find the axis of symmetry, find \(x=-\frac{b}{2a}.\)


	The axis of symmetry is \(x=3.\)

The vertex is on the line \(x=3.\)
Let \(x=3.\)


	The vertex is \((3,1).\)

The y -intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Simplify.
	The y -intercept is \((0,-8).\)
The point \((0,-8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-8).\)	Point symmetric to the y -intercept is \((6,-8).\)

The x -intercept occurs when \(y=0.\)
Let \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x .
	The x -intercepts are \((4,0),(2,0).\)
Graph the parabola.

Graph \(y=2{x}^{2}+4x+5\) .
Graph \(y=-{\left(x-3\right)}^{2}+5\) .

Graph \(x=2{(y-2)}^{2}+1\) .

Révèle la réponse


Identify the constants a, h, k .	\(a=2,\) \(h=1,\) \(k=2\)
Since \(a=2,\) the parabola opens to the right.
The axis of symmetry is \(y=k.\)	\(\\) The axis of symmetry is \(y=2.\)
The vertex is \((h,k).\)	\(\\) The vertex is \((1,2).\)
Find the x -intercept by substituting \(y=0.\)	\(\begin{array}{lll}x & = & 2{(y-2)}^{2}+1 \\ x & = & 2{(0-2)}^{2}+1 \\ x & = & 9\end{array}\)
	\(\\) The x -intercept is \((9,0).\)
Find the point symmetric to \((9,0)\) across the axis of symmetry.	\(\ (9,4)\)
Find the y -intercepts. Let \(x=0.\)	\(\begin{array}{lll}x & = & 2{(y-2)}^{2}+1 \\ 0 & = & 2{(y-2)}^{2}+1 \\ -1 & = & 2{(y-2)}^{2}\end{array}\)
	A square cannot be negative, so there is no real solution. So there are no y -intercepts.
Graph the parabola.

Graph \(x=-2{\left(y+2\right)}^{2}+4\) .
Graph \(x=-{y}^{2}+2y-3\) .
Graph \({y}^{2}=24x.\) Identify and label the focus, directrix, and endpoints of the latus rectum.
Révèle la réponse
The standard form that applies to the given equation is \({y}^{2}=4px.\) Thus, the axis of symmetry is the x-axis. It follows that:
- \(24=4p,\) so \(p=6.\) Since \(p>0,\) the parabola opens right
- the coordinates of the focus are \((p,0)=(6,0)\)
- the equation of the directrix is \(x=-p=-6\)
- the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute \(x=6\) into the original equation: \((6,\pm 12)\)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Graph \({y}^{2}=-16x.\) Identify and label the focus, directrix, and endpoints of the latus rectum.

Révèle la réponse

Focus: \((-4,0);\) Directrix: \(x=4;\) Endpoints of the latus rectum: \((-4,\pm 8)\)
Graph \({x}^{2}=-6y.\) Identify and label the focus, directrix, and endpoints of the latus rectum.
Révèle la réponse
The standard form that applies to the given equation is \({x}^{2}=4py.\) Thus, the axis of symmetry is the y-axis. It follows that:
- \(-6=4p,\) so \(p=-\frac{3}{2}.\) Since \(p<0,\) the parabola opens down.
- the coordinates of the focus are \((0,p)=(0,-\frac{3}{2})\)
- the equation of the directrix is \(y=-p=\frac{3}{2}\)
- the endpoints of the latus rectum can be found by substituting \(\ y=\frac{3}{2}\) into the original equation, \((\pm 3,-\frac{3}{2})\)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Graph \({x}^{2}=8y.\) Identify and label the focus, directrix, and endpoints of the latus rectum.

Révèle la réponse

Focus: \((0,2);\) Directrix: \(y=-2;\) Endpoints of the latus rectum: \((\pm 4,2).\)
What is the equation for the parabola with focus \((-\frac{1}{2},0)\) and directrix \(x=\frac{1}{2}?\)
Révèle la réponse
The focus has the form \((p,0),\) so the equation will have the form \({y}^{2}=4px.\)
- Multiplying \(4p,\) we have \(4p=4(-\frac{1}{2})=-2.\)
- Substituting for \(4p,\) we have \({y}^{2}=4px=-2x.\)
Therefore, the equation for the parabola is \({y}^{2}=-2x.\)
What is the equation for the parabola with focus \((0,\frac{7}{2})\) and directrix \(y=-\frac{7}{2}?\)

Révèle la réponse

\({x}^{2}=14y.\)
Graph \({(y-1)}^{2}=-16(x+3).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Révèle la réponse
The standard form that applies to the given equation is \({(y-k)}^{2}=4p(x-h).\) Thus, the axis of symmetry is parallel to the x-axis. It follows that:
- the vertex is \((h,k)=(-3,1)\)
- the axis of symmetry is \(y=k=1\)
- \(-16=4p,\) so \(p=-4.\) Since \(p<0,\) the parabola opens left.
- the coordinates of the focus are \((h+p,k)=(-3+(-4),1)=(-7,1)\)
- the equation of the directrix is \(x=h-p=-3-(-4)=1\)
- the endpoints of the latus rectum are \((h+p,k\pm 2p)=(-3+(-4),1\pm 2(-4)),\) or \((-7,-7)\) and \((-7,9)\)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See .
Graph \({(y+1)}^{2}=4(x-8).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Révèle la réponse

Vertex: \((8,-1);\) Axis of symmetry: \(y=-1;\) Focus: \((9,-1);\) Directrix: \(x=7;\) Endpoints of the latus rectum: \((9,-3)\) and \((9,1).\)
Graph \({x}^{2}-8x-28y-208=0.\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Révèle la réponse
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is \({(x-h)}^{2}=4p(y-k).\) Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable \(x\) in order to complete the square.
\[\begin{array}{l}{x}^{2}-8x-28y-208=0 \\ \ {x}^{2}-8x=28y+208 \\ \ {x}^{2}-8x+16=28y+208+16 \\ \ {(x-4)}^{2}=28y+224 \\ \ {(x-4)}^{2}=28(y+8) \\ \ {(x-4)}^{2}=4⋅7⋅(y+8)\end{array}\]
It follows that:
- the vertex is \((h,k)=(4,-8)\)
- the axis of symmetry is \(x=h=4\)
- since \(p=7,p>0\) and so the parabola opens up
- the coordinates of the focus are \((h,k+p)=(4,-8+7)=(4,-1)\)
- the equation of the directrix is \(y=k-p=-8-7=-15\)
- the endpoints of the latus rectum are \((h\pm 2p,k+p)=(4\pm 2(7),-8+7),\) or \((-10,-1)\) and \((18,-1)\)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See .
Graph \({(x+2)}^{2}=-20(y-3).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Révèle la réponse

Vertex: \((-2,3);\) Axis of symmetry: \(x=-2;\) Focus: \((-2,-2);\) Directrix: \(y=8;\) Endpoints of the latus rectum: \((-12,-2)\) and \((8,-2).\)
A cross-section of a design for a travel-sized solar fire starter is shown in . The sun’s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.
- ⓐ Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.
- ⓑ Use the equation found in part ⓐ to find the depth of the fire starter.
Révèle la réponse
- ⓐ The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form \({x}^{2}=4py,\) where \(p>0.\) The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have \(p=1.7.\) \[\begin{array}{lllll}{x}^{2}=4py & \begin{array}{llll} & & & \end{array}\text{Standard form of upward-facing parabola with vertex (0,0)} \\ {x}^{2}=4(1.7)y & \begin{array}{llll} & & & \end{array}\text{Substitute 1}\text{.7 for }p. \\ {x}^{2}=6.8y & \begin{array}{llll} & & & \end{array}\text{Multiply}.\end{array}\]
- ⓑ The dish extends \(\frac{4.5}{2}=2.25\) inches on either side of the origin. We can substitute 2.25 for \(x\) in the equation from part (a) to find the depth of the dish. \[\begin{array}{ll}\ {x}^{2}=6.8y & \text{Equation found in part (a)}. \\ {(2.25)}^{2}=6.8y & \text{Substitute 2}\text{.25 for }x. \\ \ y\approx 0.74 & \text{Solve for }y.\end{array}\]
  The dish is about 0.74 inches deep.
Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun’s rays reflect off the parabolic mirror toward the “cooker,” which is placed 320 mm from the base.
ⓐ Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the x-axis as its axis of symmetry).
ⓑ Use the equation found in part ⓐ to find the depth of the cooker.
Révèle la réponse
1. ⓐ \({y}^{2}=1280x\)
2. ⓑ The depth of the cooker is 500 mm
Define a parabola in terms of its focus and directrix.

Révèle la réponse

A parabola is the set of points in the plane that lie equidistant from a fixed point, the focus, and a fixed line, the directrix.
If the equation of a parabola is written in standard form and \(p\) is positive and the directrix is a vertical line, then what can we conclude about its graph?
If the equation of a parabola is written in standard form and \(p\) is negative and the directrix is a horizontal line, then what can we conclude about its graph?

Révèle la réponse

The graph will open down.
What is the effect on the graph of a parabola if its equation in standard form has increasing values of \(p\text{?}\)
As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?

Révèle la réponse

The distance between the focus and directrix will increase.
\({y}^{2}=4-{x}^{2}\)
\(y=4{x}^{2}\)

Révèle la réponse

yes \({x}^{2}=4(\frac{1}{16})y\)
\(3{x}^{2}-6{y}^{2}=12\)
\({(y-3)}^{2}=8(x-2)\)

Révèle la réponse

yes \({(y-3)}^{2}=4(2)(x-2)\)
\({y}^{2}+12x-6y-51=0\)
\(x=8{y}^{2}\)

Révèle la réponse

\({y}^{2}=\frac{1}{8}x,V:(0,0);F:(\frac{1}{32},0);d:x=-\frac{1}{32}\)
\(y=\frac{1}{4}{x}^{2}\)
\(y=-4{x}^{2}\)

Révèle la réponse

\({x}^{2}=-\frac{1}{4}y,V:(0,0);F:(0,-\frac{1}{16});d:y=\frac{1}{16}\)
\(x=\frac{1}{8}{y}^{2}\)
\(x=36{y}^{2}\)

Révèle la réponse

\({y}^{2}=\frac{1}{36}x,V:(0,0);F:(\frac{1}{144},0);d:x=-\frac{1}{144}\)
\(x=\frac{1}{36}{y}^{2}\)
\({(x-1)}^{2}=4(y-1)\)

Révèle la réponse

\({(x-1)}^{2}=4(y-1),V:(1,1);F:(1,2);d:y=0\)
\({(y-2)}^{2}=\frac{4}{5}(x+4)\)
\({(y-4)}^{2}=2(x+3)\)

Révèle la réponse

\({(y-4)}^{2}=2(x+3),V:(-3,4);F:(-\frac{5}{2},4);d:x=-\frac{7}{2}\)
\({(x+1)}^{2}=2(y+4)\)
\({(x+4)}^{2}=24(y+1)\)

Révèle la réponse

\({(x+4)}^{2}=24(y+1),V:(-4,-1);F:(-4,5);d:y=-7\)
\({(y+4)}^{2}=16(x+4)\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: The Parabola

Graph parabolas with vertices at the origin.
Write equations of parabolas in standard form.
Graph parabolas with vertices not at the origin.
Solve applied problems involving parabolas.
Graph vertical parabolas. (IA 11.2.1)
Graph horizontal parabolas. (IA 11.2.2)
Determine whether the parabola opens upward or downward.
Find the axis of symmetry.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Axis of Symmetry	Equation	Focus	Directrix	Endpoints of Latus Rectum
x-axis	\({y}^{2}=4px\)	\((p,\ 0)\)	\(x=-p\)	\((p,\ \pm 2p)\)
y-axis	\({x}^{2}=4py\)	\((0,\ p)\)	\(y=-p\)	\((\pm 2p,\ p)\)

Axis of Symmetry	Equation	Focus	Directrix	Endpoints of Latus Rectum
\(y=k\)	\({(y-k)}^{2}=4p(x-h)\)	\((h+p,\ k)\)	\(x=h-p\)	\((h+p,\ k\pm 2p)\)
\(x=h\)	\({(x-h)}^{2}=4p(y-k)\)	\((h,\ k+p)\)	\(y=k-p\)	\((h\pm 2p,\ k+p)\)