# 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

$y = x^2 + 5x + 10$

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.