Functions and Models Reference
Parabolas
A parabola can be studied as a quadratic graph or as a conic section. The quadratic view emphasizes vertex, intercepts, and transformations; the conic view explains focus, directrix, focal width, and the latus rectum.
Quadratic Summary Table
| Form | What It Shows | Use It For |
|---|---|---|
| $y=ax^2+bx+c$ | Standard form; y-intercept is $c$. | Axis formula, discriminant, quick y-intercept. |
| $y=a(x-h)^2+k$ | Vertex form; vertex is $(h,k)$. | Transformations, graphing, maximum or minimum. |
| $y=a(x-r_1)(x-r_2)$ | Factored form; zeros are $r_1$ and $r_2$. | X-intercepts and sign behavior. |
| $a>0$ | Opens upward. | Vertex is a minimum. |
| $a<0$ | Opens downward. | Vertex is a maximum. |
Quadratic Formulas
Axis of Symmetry
$$x=-\frac{b}{2a}$$
Vertex from Standard Form
$$h=-\frac{b}{2a},\quad k=f(h)$$
Vertex Form
$$y=a(x-h)^2+k$$
Conic Connection
$$y=a(x-h)^2+k\quad\Longleftrightarrow\quad a=\frac1{4p}$$
The value of $|a|$ controls how narrow or wide a vertical parabola looks. In conic language, that same width is measured by $p$ and the focal width $|4p|$.
Quadratic Classic Examples
Vertex and Axis
Find the vertex and axis of symmetry for $f(x)=2x^2-8x+3$.
Parabola MoveUse $x=-\frac b{2a}$, then substitute to find the y-coordinate.
$$h=-\frac{-8}{2(2)}$$
$$=2$$
$$k=f(2)$$
$$=2(2)^2-8(2)+3$$
$$=-5$$
$$\text{vertex: }(2,-5)$$
$$\text{axis: }x=2$$
Read Transformations
Describe $g(x)=-3(x+1)^2+4$.
Parabola MoveRead vertex form directly: $a(x-h)^2+k$.
$$g(x)=-3(x-(-1))^2+4$$
$$\text{vertex: }(-1,4)$$
$$\text{opens downward}$$
$$\text{vertical stretch by }3$$
$$\text{maximum value: }4$$
Conic Summary Table
| Form | Opens | Focus and Directrix |
|---|---|---|
| $(x-h)^2=4p(y-k)$ | Up if $p>0$, down if $p<0$. | Focus $(h,k+p)$, directrix $y=k-p$. |
| $(y-k)^2=4p(x-h)$ | Right if $p>0$, left if $p<0$. | Focus $(h+p,k)$, directrix $x=h-p$. |
| Vertex | Turning point. | $(h,k)$ |
| Axis of symmetry | Line through vertex and focus. | $x=h$ for vertical parabolas; $y=k$ for horizontal parabolas. |
| Latus rectum | Segment through the focus perpendicular to the axis. | Length $|4p|$. |
Conic Formulas
Vertical Axis
$$(x-h)^2=4p(y-k)$$ $$\text{focus: }(h,k+p)$$ $$\text{directrix: }y=k-p$$
Horizontal Axis
$$(y-k)^2=4p(x-h)$$ $$\text{focus: }(h+p,k)$$ $$\text{directrix: }x=h-p$$
Focal Width
$$\text{latus rectum length}=|4p|$$
Parameter
$$p=\text{directed distance from vertex to focus}$$
The squared variable tells us the direction: if $x$ is squared, the parabola opens up or down; if $y$ is squared, it opens left or right.
Conic Classic Examples
Find Focus and Directrix
Find the vertex, focus, directrix, axis, and focal width of $(x-2)^2=12(y+1)$.
Parabola MoveMatch the equation to $(x-h)^2=4p(y-k)$.
$$(x-2)^2=12(y-(-1))$$
$$h=2,\quad k=-1,\quad 4p=12$$
$$p=3$$
$$\text{vertex: }(2,-1)$$
$$\text{focus: }(2,2)$$
$$\text{directrix: }y=-4$$
$$\text{axis: }x=2$$
$$\text{focal width: }12$$
Write an Equation
A parabola has vertex $(1,-2)$ and focus $(5,-2)$. Write its equation.
Parabola MoveThe focus is to the right of the vertex, so use the horizontal form.
$$h=1,\quad k=-2,\quad p=4$$
$$(y-k)^2=4p(x-h)$$
$$(y+2)^2=4(4)(x-1)$$
$$(y+2)^2=16(x-1)$$
Latus Rectum Summary Table
| Parabola Type | Latus Rectum Line | Endpoints |
|---|---|---|
| $(x-h)^2=4p(y-k)$ | $y=k+p$ | $(h-2p,k+p)$ and $(h+2p,k+p)$ |
| $(y-k)^2=4p(x-h)$ | $x=h+p$ | $(h+p,k-2p)$ and $(h+p,k+2p)$ |
Latus Rectum Formulas
Vertical Parabola
$$\text{line: }y=k+p$$ $$(h\pm2p,k+p)$$
Horizontal Parabola
$$\text{line: }x=h+p$$ $$(h+p,k\pm2p)$$
Length
$$|4p|$$
Half-Length
$$|2p|$$
The latus rectum is sometimes misspelled as "lattice rectum." The standard term is latus rectum.
Latus Rectum Classic Examples
Vertical Latus Rectum
Find the latus rectum line and endpoints for $(x-2)^2=12(y+1)$.
Parabola MoveUse the same $h$, $k$, and $p$ from standard form. For a vertical parabola, the latus rectum line is horizontal.
$$h=2,\quad k=-1,\quad p=3$$
$$\text{line: }y=k+p$$
$$y=2$$
$$(h-2p,k+p)=(2-6,2)=(-4,2)$$
$$(h+2p,k+p)=(2+6,2)=(8,2)$$
$$\text{length: }12$$
Horizontal Latus Rectum
Find the latus rectum line and endpoints for $(y+2)^2=16(x-1)$.
Parabola MoveFor a horizontal parabola, the latus rectum line is vertical through the focus.
$$h=1,\quad k=-2,\quad 4p=16$$
$$p=4$$
$$\text{line: }x=h+p$$
$$x=5$$
$$(h+p,k-2p)=(5,-2-8)=(5,-10)$$
$$(h+p,k+2p)=(5,-2+8)=(5,6)$$
$$\text{length: }16$$
Parabola Checklist
| Step | Question |
|---|---|
| 1 | Is the parabola being treated as a function graph or a conic section? |
| 2 | Which variable is squared? |
| 3 | What are $h$, $k$, and $p$? |
| 4 | Where are the vertex, focus, directrix, and axis? |
| 5 | What is the latus rectum line and what are its endpoints? |
| 6 | Does the graph need intercepts, domain, range, or maximum/minimum value? |