Free Online Graphing Calculator

Plot functions, implicit curves, inequalities, polar and parametric graphs, and slope fields in your browser. Click any curve to see its roots, extrema and intersections, then share the graph as a link or PNG.

Try:

On this page

  1. How to use the graphing calculator
  2. The math behind the graphs
  3. Example 1: where a line meets a circle
  4. Example 2: a polar rose, drawn once
  5. Example 3: shading a feasible region
  6. Example 4: a slope field with a solution curve
  7. Reading the result and common mistakes

How to use the graphing calculator

This free online graphing calculator is for students who need a plot without buying a handheld, and for engineers who want a quick picture of an equation. It also opens as the Graph tab beside the scientific calculator.

  1. Type in the expression list above. Each row is one item and the graph redraws as you type.
  2. Click a curve. Grey dots mark its roots, maxima, minima, y-intercept and crossings with other curves. Hover for coordinates.
  3. Drag to pan, scroll or pinch to zoom. Share copies a link to your graph, and the download button saves a PNG.
To drawType
A functiony = x^2 - 4 or x^2 - 4
An implicit curvex^2 + y^2 = 25
An inequalityy < x^2, x^2 + y^2 <= 9
A polar curver = 1 + cos(θ)
A parametric curve(cos(3t), sin(2t))
A limited domainy = sqrt(x) {0 < x < 4}
A point(2, 3)
A slidera = 2, then use a elsewhere
A slope fielddy/dx = x - y
A vector field<-y, x>

The math behind the graphs

Roots and crossings are found by scanning the visible window for sign changes, so they only appear for what is on screen.

Example 1: where a line meets a circle

Where does y = x + 1 cross x² + y² = 25? Substitute: x² + (x + 1)² = 25, so 2x² + 2x − 24 = 0, with discriminant (b² − 4ac) 196 and roots x = 3 and x = −4. The points are (3, 4) and (−4, −3), and both satisfy the circle: 9 + 16 = 25 and 16 + 9 = 25. The chord between them is 7√2 ≈ 9.899495 long.

Crossing dots appear for curves written y = f(x), so draw the circle as two halves, y = sqrt(25 - x^2) and y = -sqrt(25 - x^2), and make the line y = x + a with a slider a = 1. Click the line to see both points. Then drag the slider: the discriminant becomes 200 − 4a².

aDiscriminantCrossings
11962, at x = 3 and −4
51002, at x = 0 and −5
742, at x = −3 and −4
8−56none

The line is tangent (one touching point) when the discriminant is 0, at a = 5√2 ≈ 7.071068.

Example 2: a polar rose, drawn once

Type r = cos(3θ). On the default 0 to 2π you get three petals, traced twice: the point at θ + π equals the point at θ, because cos(3θ + 3π) = −cos(3θ) and the negative r flips the point across the origin. Computed over a full sweep, the gap between the two is about 4 × 10⁻¹⁵, which is zero. One pass of length π is enough:

r = cos(3θ) {0 < θ < π}

The petal tips are at θ = 0 (r = 1, point (1, 0)), θ = 2π/3 (point (−0.5, 0.866025)) and θ = π/3, where r = −1 puts the tip at (−0.5, −0.866025). The curve returns to the origin at θ = π/6, π/2 and 5π/6.

The range changes area. Area is ½∫r² dθ: over 0 to π it is π/4 ≈ 0.7853982, the true area, and over 0 to 2π it is 1.570796, double, because the rose is counted twice. For r = cos(kθ), odd k repeats after π, but even k does not: r = cos(2θ) has four petals and needs the full 2π.

Example 3: shading a feasible region

Constraints: x ≥ 0, y ≥ 0, x + y ≤ 6, 2x + y ≤ 8. Type each as a row (use >= and <=). Each row shades its own half-plane, and the feasible region, where all four hold, is the darkest overlap.

Corners are where two boundaries meet. Subtracting x + y = 6 from 2x + y = 8 gives x = 2, y = 4. The others are (0, 0), (4, 0) and (0, 6). The grapher does not list corners, so add the points row (0, 0), (4, 0), (2, 4), (0, 6) and check each dot lands on a corner.

To maximise 4x + 3y, test the corners: 0, 16, 20 and 18. The best is 20, at (2, 4). The area of the region is 14. A spot check: (1, 1) is feasible, but (4, 3) is not, since 4 + 3 = 7 exceeds 6.

Example 4: a slope field with a solution curve

For dy/dx = x − y the slope at (0, 1) is −1, at (2, 0) it is 2, and on y = x it is 0. Solving by hand with y(0) = 1 gives y = x − 1 + 2e^(−x). Check: y′ = 1 − 2e^(−x), and x − y is the same. The grapher draws the field but does not solve it, so overlay your solution and watch it follow the segments. Values: y(1) = 2/e ≈ 0.7357589, y(2) ≈ 1.270671. Click the curve and its minimum sits at x = ln 2 ≈ 0.693147, on y = x, where the field is flat.

Reading the result and common mistakes

Frequently asked questions

Is this online graphing calculator free?

Yes. You can plot, share a link and download a PNG without paying. Nothing to install, since it runs in your browser.

Can it replace a TI-84 for plotting?

For graphing, largely yes: functions, polar, parametric, inequalities, sliders, and roots on click. It does not run TI programs, and it will not help in an exam that requires an approved device.

How do I graph inequalities and shade the region?

Type the inequality as a row, such as y < x^2 or x^2 + y^2 <= 9. The side where it holds is shaded. Add more rows to see where several regions overlap.

How do I find where two graphs intersect?

Write both as y = f(x), then click one. A grey dot appears at each crossing with the other, and hovering shows coordinates. Split an implicit curve such as a circle into upper and lower halves first.

How is this different from Desmos?

Both plot functions, inequalities and sliders in the browser. Desmos has saved accounts and classroom activities, so if your teacher uses those, Desmos is the better choice. This page is a quick plotter that sits beside a scientific calculator.

The WideMath team

Anuj Paudel, Ashish Pandey and Jhalak Prasad Pandey, Nepal

We build WideMath's free calculators. Every worked example on this page was computed with the tool above and checked against an independent method before it was published. Found a wrong answer? Tell us at [email protected].