What is a parabola?
A parabola is the graph of a quadratic function: any function you can write as
y = ax² + bx + c, with a ≠ 0
That is the standard form of the parabola equation, and the three coefficients a, b, and c completely determine the curve. Every quadratic function draws the same U-shaped, perfectly symmetric arc; the only things that change are which way it opens, how wide it is, and where it sits.
The explorer above starts at y = x² - 2x - 3. Read its screen against the equation: the curve opens upward, its turning point (the vertex) is marked at (1, -4), the dashed purple line labeled x = 1 is the axis of symmetry, the teal dots at (-1, 0) and (3, 0) are the x-intercepts (the roots), and the amber dot at (0, -3) is the y-intercept. Every one of those facts is computed live from the three slider values. The rest of this page shows how to compute them on paper, so the graph stops being a picture and becomes a set of consequences.
The quadratic function: what a, b, and c each do
| Coefficient | What it controls | On the graph |
|---|---|---|
| a | direction and width | a > 0 opens up, a < 0 opens down; a bigger absolute value of a is narrower; a = 0 is not a parabola |
| b | vertex position (together with a) | swings the vertex along a curved path; leaves the y-intercept alone |
| c | y-intercept | the parabola crosses the y-axis at exactly (0, c) |
a sets the direction and the width
The sign of a decides which way the parabola opens. Positive a opens upward, so the vertex is the lowest point (a minimum). Negative a opens downward, so the vertex is the highest point (a maximum). The explorer’s “opens” readout states this directly as you cross zero.
The size of a decides the width, and here is the part most students get backwards: a bigger absolute value of a makes the parabola narrower, not wider. The reason is that a multiplies x², so it stretches the graph vertically; with a = 3, moving one unit away from the axis lifts the curve three times as far as a = 1 does, and the arms hug the axis tightly. Values of a between 0 and 1 do the opposite and flatten the curve wide.
c pins the y-intercept
Set x = 0 in y = ax² + bx + c and the a and b terms vanish, leaving y = c. So the parabola always crosses the y-axis at exactly (0, c), which is why the explorer’s slider is labeled “c (y-intercept)”. Sliding c moves the whole curve straight up and down, and it hands you a free, zero-work plotting point for any parabola you graph by hand.
b is the subtle one
If a shapes the curve and c slides it vertically, it is tempting to assume b slides it sideways. It does not. Changing b moves the vertex along a curved path while the y-intercept stays pinned at (0, c), because the point (0, c) does not contain b at all. The horizontal position of the vertex is x = -b/(2a), so b and a together decide where the turning point sits, but the shape near the y-axis stays anchored.
How to find the vertex of a parabola
The vertex is the turning point: the minimum of an upward parabola, the maximum of a downward one. You find it in two steps:
x = -b / (2a), then substitute that x back into the equation to get y
Worked example. Find the vertex of y = x² - 4x + 3. Here a = 1, b = -4, c = 3.
x = -b / (2a) = -(-4) / (2 × 1) = 4 / 2 = 2
y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -1
The vertex is (2, -1), and since a = 1 is positive the parabola opens upward, making (2, -1) its minimum. The y-intercept is (0, 3), read straight off c with no work at all.
There is a second way to see the vertex: rewrite the equation as y = a(x - h)² + k and the vertex (h, k) can be read off directly. That rewrite is vertex form, and it gets its own lesson.
Axis of symmetry: the line x = -b/(2a)
The axis of symmetry is the vertical line through the vertex:
x = -b / (2a)
It splits the parabola into two mirror-image halves: every point on one side has a twin at the same height on the other side. For y = x² - 4x + 3 the axis is x = 2, the same x-value as the vertex, always.
The mirror property is not decoration; it does real work. The two x-intercepts of this parabola turn out to be x = 1 and x = 3 (computed in the next section), and each sits exactly 1 unit from the axis: their average is (1 + 3) / 2 = 2. Two roots always average to -b/(2a). The mirror also mass-produces plotting points: the y-intercept (0, 3) sits 2 units left of the axis, so its reflection (4, 3) must also be on the curve. Check it: (4)² - 4(4) + 3 = 16 - 16 + 3 = 3. It is.
Where -b/(2a) comes from
The formula is not a convention to be memorised. It falls out of rewriting the equation by completing the square:
- y = ax² + bx + c Start with standard form.
- y = a(x² + (b/a)x) + c Factor a out of the two x terms. It has to come out of both, which is where b/a appears.
- y = a(x² + (b/a)x + (b/2a)²) + c − a(b/2a)² Add half the x coefficient, squared, inside the bracket, and subtract the same amount outside so nothing changes.
- y = a(x + b/2a)² + c − b²/4a The bracket is now a perfect square, and a(b/2a)² tidies to b²/4a.
- y = a(x − (−b/2a))² + (c − b²/4a) Written as vertex form y = a(x − h)² + k, which names h and k.
Now read the last line. The only part of it that contains x is a square, and a square is never negative and is zero at exactly one place. So the bracket adds nothing to y when x = -b/(2a), and adds something everywhere else: something positive when a > 0, pushing the curve up from its lowest point, and something negative when a < 0, pulling it down from its highest. Either way the turning point is at
x = -b/(2a), with y = c - b²/(4a)
and that pair (h, k) is what vertex form reads off directly.
There is a second route to the same place that uses none of this. The quadratic formula gives the roots as (-b + √D)/(2a) and (-b - √D)/(2a). Add them and the √D terms cancel; halve the result and you have -b/(2a). Two roots always average to the axis, so when a parabola crosses the x-axis twice, the axis of symmetry is the midpoint of the crossings. The two derivations agree, and neither one had to assume the other.
The discriminant: crossing, touching, or missing the x-axis
How many x-intercepts does a parabola have? One number answers before you graph anything, the discriminant:
b² - 4ac
| Discriminant b² - 4ac | Real roots | The graph |
|---|---|---|
| positive | two | crosses the x-axis at two points |
| zero | one (repeated) | the vertex just touches the x-axis |
| negative | none | the parabola never reaches the x-axis |
Worked example, continued. For y = x² - 4x + 3 the discriminant is (-4)² - 4(1)(3) = 16 - 12 = 4. Positive, so there are two roots, and the quadratic formula finds them:
x = (4 ± √4) / 2 = (4 ± 2) / 2, so x = 1 and x = 3
Those are the two points where the parabola crosses the x-axis, symmetric about x = 2 exactly as the axis of symmetry promised.
Ten worked parabolas
Every column here is computed from a, b and c by the same formulas above, so the table is a set of answers to check your own working against.
| Equation | Opens | Vertex | Axis | y-intercept | b² − 4ac | x-intercepts |
|---|---|---|---|---|---|---|
| y = x² - 4x + 3 | upward | (2, -1) | x = 2 | (0, 3) | 4 | 1, 3 |
| y = x² - 2x - 3 | upward | (1, -4) | x = 1 | (0, -3) | 16 | -1, 3 |
| y = -x² + 4x - 3 | downward | (2, 1) | x = 2 | (0, -3) | 4 | 1, 3 |
| y = x² - 4x + 4 | upward | (2, 0) | x = 2 | (0, 4) | 0 | 2 |
| y = x² + 1 | upward | (0, 1) | x = 0 | (0, 1) | -4 | none |
| y = -x² - 2x - 3 | downward | (-1, -2) | x = -1 | (0, -3) | -8 | none |
| y = 2x² + 3x - 2 | upward | ≈ (-0.75, -3.13) | x = -0.75 | (0, -2) | 25 | -2, 0.5 |
| y = 0.5x² - 2x | upward | (2, -2) | x = 2 | (0, 0) | 4 | 0, 4 |
| y = 3x² - 6x + 1 | upward | (1, -2) | x = 1 | (0, 1) | 24 | ≈ 0.18, 1.82 |
| y = x² - 5x | upward | (2.5, -6.25) | x = 2.5 | (0, 0) | 25 | 0, 5 |
Read down the discriminant column against the last one and the trichotomy is plain: positive gives two crossings, zero gives one, negative gives none. Note the marked approximation on 3x² − 6x + 1: its roots are (3 ± √6)/3, and no two-decimal number is either of them.
Which ways a parabola can open
Every parabola on this page so far has opened up or down, because y = ax² + bx + c can only do those two things. Turn the equation on its side and two more appear.
| Shape | Equation | A function of x? | Notes |
|---|---|---|---|
| Opens upward | y = ax² + bx + c, a > 0 | yes | The vertex is the lowest point, a minimum. This is the default parabola. |
| Opens downward | y = ax² + bx + c, a < 0 | yes | The same curve flipped. The vertex is now the highest point, a maximum. |
| Opens right | x = ay² + by + c, a > 0 | no | A sideways parabola. It fails the vertical line test, so it is a relation but not a function of x. |
| Opens left | x = ay² + by + c, a < 0 | no | The mirror of the one above, and equally not a function of x. |
| Half a parabola | y = √x | yes | Square-rooting keeps only the non-negative branch, so this is the upper half of the sideways parabola x = y². Half of it is a function; the whole of it is not. |
The sideways cases matter for a reason worth being clear about: they are parabolas, but they are not functions of x. A vertical line drawn through a sideways parabola hits it twice, so a single x has two y values, and that is exactly what a function is not allowed to do. The usual repair is to take one branch, which is what the square root does: y = √x is the top half of x = y², and half of a sideways parabola is a perfectly good function.
How to graph a parabola, step by step
Graphing quadratic functions by hand comes down to extracting five facts from a, b, and c, in this order:
- Direction. Read the sign of a: positive opens up, negative opens down. Note roughly how wide from the size of a (big means narrow).
- Axis and vertex. Compute x = -b/(2a) for the axis of symmetry, substitute that x back in for y, and plot the vertex.
- Intercepts. Plot the y-intercept (0, c). Compute the discriminant b² - 4ac: if it is positive or zero, find the x-intercept(s) by factoring or the quadratic formula and plot them too.
- Mirror points. Reflect the y-intercept (and any other point you have) across the axis of symmetry for free extra points.
- Sweep the curve. Draw one smooth U through the points, both arms, symmetric about the axis. No straight segments and no sharp corner at the vertex.
Run the recipe on the explorer’s starting equation, y = x² - 2x - 3: a = 1 opens up; axis x = -(-2)/(2 × 1) = 1; vertex y = 1 - 2 - 3 = -4, so (1, -4); y-intercept (0, -3), which mirrors to (2, -3); discriminant 4 + 12 = 16, so roots x = (2 ± 4)/2 = -1 and 3. Hit Reset on the explorer and check all of it against the screen.
The explorer above also works as a quadratic graphing calculator: set a, b, and c and read the vertex, axis, discriminant, and intercepts at a glance.
It’s free to embed on your own site or LMS. Next, read the vertex straight off the equation in vertex form, or drop back to the a = 0 case and master lines in slope-intercept form.