# 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 [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the vertex will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and [latex]y[/latex] replaced by [latex]\left(y-k\right)[/latex].
To graph parabolas with a vertex [latex]\left(h,k\right)[/latex] other than the origin, we use the standard form [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex] for parabolas that have an axis of symmetry parallel to the *x*-axis, and [latex]{\left(x-h\right)}^{2}=4p\left(y-k\right)[/latex] 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.

### A General Note: Standard Forms of Parabolas with Vertex (*h*, *k*)

The table and Figure 9 summarize the standard features of parabolas with a vertex at a point [latex]\left(h,k\right)[/latex].
Axis of Symmetry |
Equation |
Focus |
Directrix |
Endpoints of Latus Rectum |

[latex]y=k[/latex] | [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex] | [latex]\left(h+p,\text{ }k\right)[/latex] | [latex]x=h-p[/latex] | [latex]\left(h+p,\text{ }k\pm 2p\right)[/latex] |

[latex]x=h[/latex] | [latex]{\left(x-h\right)}^{2}=4p\left(y-k\right)[/latex] | [latex]\left(h,\text{ }k+p\right)[/latex] | [latex]y=k-p[/latex] | [latex]\left(h\pm 2p,\text{ }k+p\right)[/latex] |

**Figure 9.**(a) When [latex]p>0[/latex], the parabola opens right. (b) When [latex]p<0[/latex], the parabola opens left. (c) When [latex]p>0[/latex], the parabola opens up. (d) When [latex]p<0[/latex], the parabola opens down.

### How To: Given a standard form equation for a parabola centered at (*h*, *k*), sketch the graph.

- Determine which of the standard forms applies to the given equation: [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex] or [latex]{\left(x-h\right)}^{2}=4p\left(y-k\right)[/latex].
- Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
- If the equation is in the form [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex], then:
- use the given equation to identify [latex]h[/latex] and [latex]k[/latex] for the vertex, [latex]\left(h,k\right)[/latex]
- use the value of [latex]k[/latex] to determine the axis of symmetry, [latex]y=k[/latex]
- set [latex]4p[/latex] equal to the coefficient of [latex]\left(x-h\right)[/latex] in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens right. If [latex]p<0[/latex], the parabola opens left.
- use [latex]h,k[/latex], and [latex]p[/latex] to find the coordinates of the focus, [latex]\left(h+p,\text{ }k\right)[/latex]
- use [latex]h[/latex] and [latex]p[/latex] to find the equation of the directrix, [latex]x=h-p[/latex]
- use [latex]h,k[/latex], and [latex]p[/latex] to find the endpoints of the latus rectum, [latex]\left(h+p,k\pm 2p\right)[/latex]

- If the equation is in the form [latex]{\left(x-h\right)}^{2}=4p\left(y-k\right)[/latex], then:
- use the given equation to identify [latex]h[/latex] and [latex]k[/latex] for the vertex, [latex]\left(h,k\right)[/latex]
- use the value of [latex]h[/latex] to determine the axis of symmetry, [latex]x=h[/latex]
- set [latex]4p[/latex] equal to the coefficient of [latex]\left(y-k\right)[/latex] in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens up. If [latex]p<0[/latex], the parabola opens down.
- use [latex]h,k[/latex], and [latex]p[/latex] to find the coordinates of the focus, [latex]\left(h,\text{ }k+p\right)[/latex]
- use [latex]k[/latex] and [latex]p[/latex] to find the equation of the directrix, [latex]y=k-p[/latex]
- use [latex]h,k[/latex], and [latex]p[/latex] to find the endpoints of the latus rectum, [latex]\left(h\pm 2p,\text{ }k+p\right)[/latex]

- If the equation is in the form [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex], then:
- Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

### Example 5: Graphing a Parabola with Vertex (*h*, *k*) and Axis of Symmetry Parallel to the *x*-axis

Graph [latex]{\left(y - 1\right)}^{2}=-16\left(x+3\right)[/latex]. 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 [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex]. Thus, the axis of symmetry is parallel to the*x*-axis. It follows that:

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

**Figure 10**

### Try It 5

Graph [latex]{\left(y+1\right)}^{2}=4\left(x - 8\right)[/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution### Example 6: Graphing a Parabola from an Equation Given in General Form

Graph [latex]{x}^{2}-8x - 28y - 208=0[/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.### Solution

Start by writing the equation of the**parabola**in standard form. The standard form that applies to the given equation is [latex]{\left(x-h\right)}^{2}=4p\left(y-k\right)[/latex]. 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 [latex]x[/latex] in order to complete the square.

- the vertex is [latex]\left(h,k\right)=\left(4,-8\right)[/latex]
- the axis of symmetry is [latex]x=h=4[/latex]
- since [latex]p=7,p>0[/latex] and so the parabola opens up
- the coordinates of the focus are [latex]\left(h,k+p\right)=\left(4,-8+7\right)=\left(4,-1\right)[/latex]
- the equation of the directrix is [latex]y=k-p=-8 - 7=-15[/latex]
- the endpoints of the latus rectum are [latex]\left(h\pm 2p,k+p\right)=\left(4\pm 2\left(7\right),-8+7\right)[/latex], or [latex]\left(-10,-1\right)[/latex] and [latex]\left(18,-1\right)[/latex]

**Figure 11**

### Try It 6

Graph [latex]{\left(x+2\right)}^{2}=-20\left(y - 3\right)[/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution## Licenses & Attributions

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- Precalculus.
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