# Function Calculator + Online Solver With Free Steps

The **Function Calculator **calculates the properties of a graph drawn from the equation entered in the calculator. Furthermore, the graph is drawn on a cartesian plane to represent that function **f(x)** visually**.**

This calculator takes in any kind of legitimate equation and will detect the** type** of the function (linear line, conic shape, etc.) and then provide a table showing its **properties** based on it.

There could be two graphs, with one showing a **zoomed-in view** and the other expressing the **overall trend** of the function given on the graph.

Moreover, this calculator supports **three-dimensional equations**, except that it does not give a detailed properties table for these figures and does not draw a graph to express them visually. For **higher-order polynomials**, the results show only the visual representation of the functions, and their properties are **unavailable**.

## What Is the Function Calculator?

**The Function Calculator determines the type of function and its properties and displays it based on the inputted polynomial in the calculator text box. Also, it draws a graph to visually represent the trend of the line on the cartesian plane. The properties differ from function to function based on the type of shape it expresses. **

The calculator consists of a single-line text box labeled “**Enter the Function, y = f(x) =,**” where you can enter any legitimate function to find its properties and type. Additionally, it is recommended to use the calculator for two-dimensional functions as it correctly represents its properties in those functions.

## How To Use the Function Calculator?

You can correctly use the **Function Calculator **by entering the desired function into the text box and submitting it. A new window will open, showing all the properties and graphs for the respective function **f(x)**.

The stepwise guidelines for the calculator’s usage are below:

### Step 1

Enter the desired equation into the** text box**.

### Step 2

Ensure that the function is written **correctly**, following the **latex format** for the calculator to calculate.

### Step 3

Press the “**Submit**” button to get the results.

### Results

A pop-up window appears showing the detailed results in the sections explained below:

**Input:**This section shows the input function written in the**mathematical equation**. You can**verify**the equation by matching your entered function with the input interpreted by the calculator and edit accordingly.**Geometric Figure:**This section shows the type of function and its geometric name. Based on its geometric shape, it also shows its properties such as y-intercept, gradient, directrix, focus, vertex, curvature, etc. You can enable the properties by clicking on the “**Show Properties**” button on the top right-hand of the section.**Plots:**Here, the graph is drawn for the functions on the**cartesian plane**. The linear line plots will have only one plot with a specific range of x-values. Furthermore, the conics are**plotted twice**, with one showing the**zoomed-in**section and the other showing the**overall trend**of the graph at a wider range of x-values.

## Solved Examples

### Example 1

Suppose a **quadratic equation**:

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

Find the** type of conic**, which the functions represent, and calculate its **respective properties**.

### Solution

As we know, this type of quadratic equation is usually considered a **parabola**, which can be converted into the **standard parabola form**:

\[ y = a(x\,–\, h)^2 + k \]

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

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

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

After converting to the standard form, we can find the properties of the parabola by simply comparing it to the **standard form equation**:

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

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

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 the 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 **linear line** with the equation:

\[ y = 4x + 7 \]\

Find the **properties** of the line expressed by this function.

### Solution

A linear line equation is governed by the **general linear equation**:

\[ y = mx + c \]

Where **y** is the **y-coordinate**, **x** is the **x-coordinate**, **m** is the **gradient/slope** of the line, and **c **is the **y-intercept**, which is the y-coordinate at which the line **intersects** the** y-axis.**

Comparing the equation given with the general line equation, we can write the **properties** of the line as:

\[\text{y-intercept = c = } 7 \]

For the **x-intercept**, we find the value of x-coordinate when $y = 0$. Hence:

\[ 0 = 4x + 7 \]

\[ -4x = 7 \]

\[ \mathbf{x = -\frac{7}{4}} \]

The **gradient** for this is expressed as **m**, which is:

\[ \text{Gradient = m = } 4 \]

To find the **Normal Vector** of this line, we need to write the above equation in the form of $ax + by = 0$

\[ y = 4x + 7 \]

\[ -4x + x = 7 \]

Neglecting the constant value, we can take the** coefficients**, a and b, of the x and y variables as the normal coordinates (-4, 1). Now we will divide both the coordinates with the** magnitude** of the point that is described as:

\[ \text{Magnitude = } \sqrt{x^2 + y^2}\]

\[ \text{Magnitude = } \sqrt{(-4)^2 + 1^2}\]

\[ \text{Magnitude = } \sqrt{17} \]

Thus, the normal vector coordinates are $\left(-\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right)$. They are approximated in the** decimal form** as (-0.970, 0.242).

The above line is a **linear equation**. Hence, its **curvature** is equal to zero.