Find the x-intercept where your graph crosses. Use slope-intercept, standard, or quadratic inputs with steps. Compare methods, check domain rules, and export results quickly.
| Type | Inputs | Expected x-intercept(s) |
|---|---|---|
| Line (y = mx + b) | m = 2, b = −8 | x = 4 |
| Standard (Ax + By + C = 0) | A = 3, B = −2, C = 12 | x = −4 |
| Quadratic (ax² + bx + c) | a = 1, b = −5, c = 6 | x = 2 and x = 3 |
| Circle | h = 2, k = 1, r = 5 | x = 2 ± √24 |
| Exponential | a = 3, b = 2, c = −12 | x = 2 |
It is the x-value where a graph crosses the x-axis. At that point, the y-value equals zero, so you solve the equation with y set to 0.
Yes. Quadratics can have two, one, or zero real x-intercepts. Higher-degree polynomials can have several, depending on their shape and real roots.
Some equations never reach y = 0. For example, a circle may sit above the x-axis, or a quadratic can have a negative discriminant, producing complex roots.
If a horizontal line is y = k and k ≠ 0, it never touches the x-axis, so there is no x-intercept. If k = 0, every x is an intercept.
A vertical line has the form x = constant. It crosses the x-axis at (constant, 0), so the x-intercept equals that constant value.
For y = a·b^x + c, you need −c/a > 0 because b^x is always positive. If the target is zero or negative, no real x-intercept exists.
Logarithms require a positive argument. For log_b(x − h), you must have x > h. The calculator solves the equation and keeps this domain in mind.
Polynomial roots are approximated by scanning for sign changes and refining with bisection. Use a smaller step size and a wider range for better detection.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.