A quadratic equation is any equation where the highest power of the variable is 2, written in the general form ax² + bx + c = 0. The values a, b, and c are real numbers, and a cannot equal zero (if it did, the x² term would disappear and the equation would just be linear). Writing one comes down to identifying which form fits the information you have, then plugging in the right values.
The Three Forms of a Quadratic Equation
Quadratic equations can be written in three different ways, and each one is useful in different situations. The form you choose depends on what information you’re starting with.
General form: f(x) = ax² + bx + c. This is the most common version you’ll see in textbooks. The coefficient “a” controls whether the parabola opens upward (positive) or downward (negative) and how wide or narrow it is. The value “c” is the y-intercept, the point where the curve crosses the vertical axis.
Vertex form: f(x) = a(x − h)² + k. Here, (h, k) is the vertex of the parabola, its highest or lowest point. The value h tells you how far the graph has shifted left or right from the origin, and k tells you how far it has shifted up or down. If someone gives you the vertex and one other point, this is the form to use.
Factored form: f(x) = a(x − r₁)(x − r₂). The values r₁ and r₂ are the x-intercepts, the points where the parabola crosses the horizontal axis. If you know where the curve hits the x-axis, this form lets you write the equation almost immediately.
Writing an Equation in General Form
If you’re given three points on a parabola, general form is your starting point. Substitute each point’s x and y values into ax² + bx + c = y, which gives you three equations with three unknowns (a, b, and c). Solve that system and you have your equation.
For example, suppose a parabola passes through (0, −10), (1, −12), and (5, 0). Plugging in (0, −10) gives you c = −10 right away, since both the ax² and bx terms vanish when x is zero. Plugging in the other two points gives you two equations with just a and b, which you can solve by substitution or elimination. In this case, you’d get a = 1, b = −3, and c = −10, so the equation is x² − 3x − 10 = 0.
Writing an Equation from the Vertex
When you can see the vertex on a graph, or someone gives it to you directly, vertex form saves a lot of work. Start with f(x) = a(x − h)² + k, plug in h and k from the vertex, then use one additional point on the curve to solve for a.
Say the vertex is at (3, −4) and the parabola also passes through (5, 0). Your equation starts as f(x) = a(x − 3)² − 4. Substituting the point (5, 0) gives you 0 = a(5 − 3)² − 4, which simplifies to 0 = 4a − 4, so a = 1. The final equation is f(x) = (x − 3)² − 4. If you need it in general form, expand the squared term and combine like terms to get f(x) = x² − 6x + 5.
Writing an Equation from X-Intercepts
If you know where the parabola crosses the x-axis, factored form is the fastest route. The x-intercepts plug directly into f(x) = a(x − r₁)(x − r₂). You still need one more point (any point on the curve, including the y-intercept) to find a.
Take x-intercepts at x = −2 and x = 5. The equation starts as f(x) = a(x + 2)(x − 5). If the curve also passes through (0, −10), substitute those values: −10 = a(0 + 2)(0 − 5), which gives −10 = −10a, so a = 1. The equation is f(x) = (x + 2)(x − 5). Expanding this produces f(x) = x² − 3x − 10, matching the general form from the earlier example. The same parabola, written three different ways depending on what information you started with.
Converting Between Forms
Going from factored or vertex form to general form is straightforward: just expand the parentheses and combine like terms. Going the other direction, from general form to vertex form, requires a technique called completing the square.
Here’s the process for converting x² − 6x + 5 into vertex form. First, group the x terms: (x² − 6x) + 5. Take half of the coefficient on x (half of −6 is −3), then square it (9). Add and subtract that value inside the expression: (x² − 6x + 9 − 9) + 5. The first three terms form a perfect square: (x − 3)² − 9 + 5, which simplifies to (x − 3)² − 4. Now you can read the vertex directly as (3, −4).
If the leading coefficient isn’t 1, divide the entire equation by that coefficient first. For instance, to complete the square on 2x² + 8x + 6, factor out the 2 to get 2(x² + 4x) + 6, then complete the square inside the parentheses.
How the Discriminant Tells You What to Expect
Before you solve or graph a quadratic equation, you can check a single value to know how many solutions it has. The discriminant is the expression b² − 4ac, pulled from the general form ax² + bx + c = 0.
If b² − 4ac is positive, the equation has two distinct real solutions, meaning the parabola crosses the x-axis in two places. If it equals exactly zero, there’s one repeated solution, and the parabola just touches the x-axis at its vertex. If it’s negative, there are no real solutions, and the parabola never reaches the x-axis at all.
This is useful when writing equations to match specific conditions. If someone asks you to write a quadratic with no real roots, you need to choose a, b, and c so that b² − 4ac comes out negative. For example, x² + 2x + 5 = 0 has a discriminant of 4 − 20 = −16, confirming it has no real solutions.
Writing Quadratics for Real-World Problems
Quadratic equations show up naturally whenever something accelerates. The most common example is projectile motion. When you throw a ball upward, its height over time follows a quadratic pattern because gravity constantly accelerates it downward.
The general physics model for vertical position is y = v₀t − ½gt², where v₀ is the initial upward speed, g is gravitational acceleration (about 10 m/s²), and t is time in seconds. If you launch a ball upward at 25 m/s from a height of 30 meters, the height equation becomes y = 30 + 25t − 5t². Setting y to zero and solving tells you when the ball hits the ground.
Business applications work similarly. If revenue depends on price and the relationship curves (as it often does), you can model it as a quadratic. Collect three data points, plug them into general form, and solve for a, b, and c.
Choosing the Right Starting Form
The key skill isn’t memorizing all three forms. It’s recognizing which one matches the information in front of you. If you can see or are given the vertex, start with vertex form. If you know the x-intercepts, start with factored form. If you have a set of points with no obvious special features, use general form and set up a system of equations.
In every case, you need the value of “a” to pin down the exact equation. Two parabolas can share the same vertex or the same x-intercepts but have completely different shapes, depending on whether “a” is large or small, positive or negative. One additional point on the curve is always enough to solve for it.