# Vertex Form Calculator + Online Solver With Free Steps

The **Vertex Form Calculator **calculates a parabolic equation’s parabolic properties in its vertex form. Furthermore, it gives the plot of the entered curve in a separate window to represent the equation visually. A Parabola is a U-shaped curve equidistant to a **focal point** and a **directrix** of the curve at any point on the parabola.

The calculator works for 2D parabolas and does not support 3D parabolic shapes such as paraboloids and cylinders. Using the equations such as $y^2 = 4ax$ in the calculator input will give the parabolic parameters, but it does not represent the plot of the equation. The calculator gives plots for quadratic or vertex form equations such as $y = a(x\,–\, h)^2 + k$

## What Is the Vertex Form Calculator?

**The Vertex Form Calculator is an online calculator that determines the properties of a parabolic equation (focus, vertex, semi-axis length, eccentricity, focal parameter, and directrix) which is in the vertex form. On top of that, it also draws the parabola’s plot under a separate heading on the window.**

The calculator interface has a single text box for inputting the parabolic equation, which is labeled “**Enter the equation of the parabola.**” You only need to enter the parabola equation in the vertex form in this single-line text box to find its parabolic properties and plots.

## How To Use the Vertex Form Calculator?

You can just enter the equation of the parabola in the text box and acquire the parabolic properties and plots to the parabola equation. Let us take a case for a parabolic equation given as follows:

\[ y = 3 (x – 6)^2 + 4 \]

You can find the properties for the above parabola equation by following the steps below:

### Step 1

Ensure that the equation of the parabola is correct and is in either vertex form or a quadratic form. In our case, it is in vertex form.

### Step 2

Input your desired parabolic equation into the single-line text box. In our situation, we type the equation as “y = 3 (x – 6)^2 + 4.” You can also enter constants and standard functions in the equation such as “**π**,” **absolute**, etc.

### Step 3

Click the **Submit **button or press the **Enter **button on the keyboard to get the results.

### Results

**Input:**This is the input section as interpreted by the calculator in LaTeX syntax. You can verify the correct interpretation of your input equation by the calculator.**Geometric Figure:**This section presents the values of the parabolic properties. The values of**focus**,**vertex**,**semi-axis length**,**eccentricity**,**focal parameter**, and**directrix**are shown. You can hide these properties by pressing the “**hide properties**” button on the top-right part of the section.**Plots:**Here, two 2D plots of parabolas are shown. The two graphs differ in perspective such that the first graph shows a closer inspection to clearly show the vertex point, whereas the second plot shows a zoomed-out view of the curve to show how the parabola curve tends to open up.

## How Does The Vertex Form Calculator Work?

The **Vertex Form Calculator **works by determining the values of the parabola equation by converting a given equation to a vertex form. To find the parabolic properties, we then compare that equation with the generalized parabola equation.

For plotting, the calculator finds the y-parameter values for a range of values of x (for a y-symmetric parabola) or vice-versa (for an x-symmetric parabola and draws a smooth curve on the plot.

### Definition

The standard quadratic form is $y = ax^2 + bx + c$, but the vertex form of the quadratic equation is $y = a(x − h)^2 + k$. In both forms, y is the y-coordinate, x is the x-coordinate, and a is a constant indicating whether the parabola points up (+a) or down (-a).

The difference between the standard form of the parabola and the vertex form is that the equation’s vertex form also gives the parabola’s vertices (h,k).

### Properties of a Parabola

To understand the workings of the calculator better, we need to understand the basic foundations of a parabola in detail. Hence, the following gives us a concise meaning of the properties:

**Axis of Symmetry (AoS):**A line that bisects the parabola into two symmetrical halves. It goes through the vertex is parallel to either x or y-axis, depending on the orientation of the parabola**Vertex:**It is the maximum (if the parabola opens downwards) or the minimum (if the parabola opens upwards) point of a parabola. In technical terms, it is a point where the derivative of a parabola is zero.**Directrix:**It is the line that is perpendicular to the AoS so that any point on the parabola is specifically equidistant from it and the focus point. This line does not intersect with the parabola.**Focus:**It is the point alongside the AoS such that any point on the parabola is equidistant from the focus and the directrix. The focus point does not lie on either the parabola or the directrix.**Semi-axis Length:**Also known as the**focal length**, it is the distance of the focus to the vertex. In parabolas, it is also equal to the distance between the parabola curve and the directrix. Hence, it is half the length of the focal parameter**Focal Parameter: The “semi-latus rectum”**is the distance between the focus and its respective directrix. For the case of parabolas, it is double the semi-axis/focal length.**Eccentricity:**This is the ratio of the distance between the vertex and focus to the distance between the vertex and directrix. The value of the eccentricity determines the conic type(hyperbola, ellipse, parabola, etc.). In the case of a parabola, the eccentricity is always equal to 1.

### Standard Vertex Form Equations

The easiest equations of parabolas to interpret are the standard vertex forms:

\[ y = a(x-h)^2 + k \tag*{(y-symmetric parabola)} \]

\[ x = a(y-k)^2 + h \tag*{(x-symmetric parabola)} \]

## Solved Examples

### Example 1

Suppose a quadratic equation:

\[ y = x^2 + 5x + 10 \]

The above equation represents a parabola. Find the focus, directrix, and length of the semi-latus rectum for **y**.

### Solution

Firstly, we convert the quadratic function into the standard vertex form of a parabola equation. By completing the square:

\[ y = x^2 + 2(1)\left(\frac{5}{2}\right)x + \frac{25}{4} + 10\, -\, \frac{25}{4}\]

\[ y = \left( x + \frac{5}{2} \right)^2 + \frac{15}{4} \]

After converting to the vertex form, we can find the properties of the parabola by simply comparing it to the generalized vector form equation:

\[ y = a(x-h)^2 + k \]

\[ \Rightarrow a > 0 = 1, h= -\frac{5}{2}, k = \frac{15}{4} \]

\[ \text{vertex} = (h,\, k) = (-\frac{5}{2},\, \frac{15}{4}) \]

The Axis of Symmetry is parallel to the y-axis and the parabola opens upwards as a > 0. Thus the semi-axis/focal length is found by:

\[ f = \frac{1}{4a} = \frac{1}{4} \]

\[ \text{Focus :} \,\, \left(\frac{5}{2},\, \frac{15}{4} + f\right) = \left(\mathbf{\frac{5}{2},\, 4}\right) \]

The directrix is perpendicular to Axis of Symmetry and hence a horizontal line:

\[ \text{Directrix :} \,\, y = \frac{15}{4}-f = \mathbf{\frac{7}{2}} \]

The length of the semi-latus rectum equals the focal parameter:

\[ \text{Focal Parameter :} \,\, p = 2f = \mathbf{\frac{1}{2}} \]

### Example 2

Consider a Vertex form equation:

\[ y = (x-12)^2 + 13 \]

Given that the vertex form equation represents a parabola. Find the focus, directrix, and length of the semi-latus rectum for **y**.

### Solution

As the vertex form is already given, we can find the parabolic properties by comparing it to the generalized vector form equation:

\[ y = a(x-h)^2 + k \]

**$\Rightarrow$ a > 0 = 1, h= 12, k = 13 **

**vertex = (h, k) = (12, 13) **

The Axis of Symmetry is parallel to the y-axis and the parabola opens upwards as a > 0. Thus the semi-axis/focal length is found by:

\[ f = \frac{1}{4a} = \frac{1}{4} \]

\[ \text{Focus :} \,\, \left(12,\, 13 + f\right) = \left(\mathbf{12,\, \frac{53}{4}}\right) \]

The directrix is perpendicular to Axis of Symmetry and hence a horizontal line:

\[ \text{Directrix :} \,\, y = -13-f = \mathbf{\frac{51}{4}} \]

The length of the semi-latus rectum equals the focal parameter:

\[ \text{Focal Parameter :} \,\, p = 2f = \mathbf{\frac{1}{2}} \]

### Example 3

Consider a Vertex form equation:

\[ x = -2(y-20)^2 + 25 \]

Given that the vertex form equation represents a parabola. Find the focus, directrix, and length of the semi-latus rectum for **x**.

### Solution

We have an equation of a parabola that is x-symmetric. Hence, we can find the parabolic properties by comparing the equation to the generalized vector form equation:

\[ x = a(y-k)^2 + h \]

**$\Rightarrow$ a < 0 = -2, h = 25 , k = 20 **

**vertex = (h, k) = (25, 20) **

The Axis of Symmetry is parallel to the y-axis, and the parabola opens to the right as a < 0. Thus the semi-axis/focal length is found by:

\[ f = \frac{1}{4a} = -\frac{1}{8} \]

\[ \text{Focus :} \,\, \left(25 + f,\, 20\right) = \left(\mathbf{\frac{199}{8},\, 20}\right) \]

The directrix is perpendicular to Axis of Symmetry and hence a horizontal line:

\[ \text{Directrix :} \,\, x = 25 – f = \mathbf{\frac{201}{8}} \]

The length of the semi-latus rectum equals the focal parameter:

\[ \text{Focal Parameter :} \,\, p = 2f = -\mathbf{\frac{1}{4}} \]