➕ Algebra 1 · Quadratics

Graphing Quadratic Functions

Graph y = ax² + bx + c: find which way the parabola opens, its vertex, axis of symmetry and intercepts, then sketch it and read its maximum or minimum.

Grades 9, 10⏱ 16 min✔ AI-checkedCCSS F-IF.C.7aCCSS F-IF.B.4CCSS A-APR.B.3

About standards and checking

Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Two blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-08.

F-IF.C.7a: Graph linear and quadratic functions and show intercepts, maxima, and minima.
F-IF.B.4: For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.
A-APR.B.3: Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

By the end you can…

Recognize a quadratic function in standard form y = ax² + bx + c (with a ≠ 0), and tell from a which way its parabola opens and whether it is narrow or wide.
Find the axis of symmetry x = −b/(2a), the vertex, the y-intercept and, when they exist, the x-intercepts.
Build a table of values around the vertex, use symmetry to plot matching points, and sketch a smooth parabola.
Read the maximum or minimum value, the domain and the range from a parabola, and explain them in a simplified real-world model.

Lesson (school style)

A quadratic function is a function whose rule can be written y = ax² + bx + c, where a, b and c are numbers and a ≠ 0. Its graph is a smooth, U-shaped curve called a parabola. In the lesson on graphing linear equations you needed only two points to draw a line. A parabola bends, so you need to know where it turns, which way it opens and where it crosses the axes. This lesson shows how to read all of that from the equation, and how to sketch the curve quickly by using its symmetry.

Quadratic functions and parabolas

A quadratic function in standard form looks like y = ax² + bx + c. The letters a, b and c stand for fixed numbers. Each number keeps its sign. For y = 2x² − 3x + 8, a = 2, b = −3 and c = 8.

The number a cannot be 0. But b or c (or both) can be 0: in y = x² − 9, b = 0, and in y = 5x², b = 0 and c = 0. If a were 0, the x² term would vanish and you would get y = bx + c, which is a line, not a parabola.

The graph of every quadratic function is a parabola. To graph one, you do not need to plot dozens of points. A few key features give you the whole picture: the direction it opens, the vertex, the axis of symmetry and the intercepts.

Which way it opens, and how wide

The sign of a decides the direction.

If a > 0, the parabola opens up, like a U.
If a < 0, the parabola opens down, like an upside-down U.

Only a decides this. The signs of b and c do not.

The size of a decides how wide the parabola is. Compare y = x², y = 2x² and y = (1/2)x². At x = 1 their y-values are 1, 2 and 1/2. For the same x (other than 0), a larger |a| puts the point farther from the x-axis, so the curve climbs faster and looks narrower. A smaller |a|, closer to 0, makes it wider. Use the absolute value |a|: y = −3x² is exactly as narrow as y = 3x², just flipped upside down.

In fact the shape of y = ax² + bx + c depends only on a. Changing b or c slides the parabola to a new position without changing its shape.

Symmetry, the axis and the vertex

A parabola is symmetric: if you fold the graph along a certain vertical line, the two halves match exactly. That line is the axis of symmetry. The point where the axis meets the parabola is the vertex, the turning point of the curve.

For y = ax² + bx + c, the axis of symmetry is the vertical line
x = −b/(2a).
In words: take the opposite of b, and divide by twice a. The vertex lies on the axis, so its x-coordinate is −b/(2a). To get its y-coordinate, substitute that x into the equation.

For y = 2x² − 16x + 35: a = 2 and b = −16, so x = −(−16)/(2 · 2) = 16/4 = 4. Then y = 2(4)² − 16(4) + 35 = 32 − 64 + 35 = 3. The axis is x = 4 and the vertex is (4, 3).

The vertex is the lowest point if the parabola opens up (a > 0) and the highest point if it opens down (a < 0).

The y-intercept and its mirror point

Every point on the y-axis has x = 0. Substitute x = 0 into y = ax² + bx + c and you get y = c. So the y-intercept is c, and the graph crosses the y-axis at the point (0, c).

Symmetry gives you a second point for free. If the axis is x = h, a point at x has a mirror image at 2h − x with the same y-value. In particular, the mirror image of (0, c) is (2h, c). If h = 0 (which happens when b = 0), the point (0, c) sits on the axis, so it is the vertex itself and has no separate mirror image.

x-intercepts (zeros)

An x-intercept is an x-value where the graph meets the x-axis. If p is an x-intercept, the graph passes through the point (p, 0). Every point on the x-axis has y = 0, so to find the x-intercepts, set y = 0 and solve. These x-values are also called the zeros of the function.

A parabola can have two, one or no x-intercepts. The vertex and the direction tell you which:

Two: the parabola opens toward the x-axis from its vertex (it opens up with the vertex below the axis, or opens down with the vertex above it).
One: the vertex is on the x-axis, so the parabola touches the axis at that point.
None: the parabola opens away from the x-axis (opens up with the vertex above it, or opens down with the vertex below it).

When the quadratic factors, you can find the x-intercepts exactly: set y = 0, factor, and set each factor equal to 0 (this uses the fact that if a product is 0, one of its factors must be 0; the lesson Solving Quadratics by Factoring explains why). When there are two x-intercepts, the axis of symmetry is exactly halfway between them. When the quadratic does not factor with integers, you can still graph it from the vertex and mirror points. Exact x-intercepts then need methods from later lessons (completing the square and the quadratic formula).

A table of values around the vertex

Find the vertex first, then build the table around it.

Put the axis x-value in the middle of your table.
Choose x-values the same distance to the left and right of the axis. For the axis x = 4, use 3 and 5, then 2 and 6, then 1 and 7.
Calculate y for one side only. A point that is the same distance from the axis, on the other side, has the same y-value.

If the axis falls between whole numbers, such as x = 3/2, choose pairs that are equally far from it: 1 and 2, then 0 and 3. The mirror trick works only for x-values that are equally far from the axis.

Plot and draw a smooth curve

Plot the vertex, the y-intercept and its mirror point, any x-intercepts, and the other points from your table. Then draw one smooth curve through them, with a rounded turn at the vertex. It is not two straight lines meeting at a sharp corner, and it is not a chain of short straight segments. A parabola keeps going, so add arrowheads at both ends and label the vertex and the axis.

Maximum, minimum, domain and range

The vertex is the lowest point of a parabola that opens up and the highest point of one that opens down. Call the vertex (h, k).

If a > 0: the minimum value of y is k, and it occurs at x = h. There is no maximum, because the graph rises forever.
If a < 0: the maximum value of y is k, and it occurs at x = h. There is no minimum, because the graph falls forever.

The maximum or minimum value is one number, a y-value. The vertex is a point. Read left to right, a parabola that opens up is decreasing until the vertex and increasing after it. One that opens down is increasing until the vertex and decreasing after it.

The domain is the set of all x-values you are allowed to use. The range is the set of all y-values that come out. For y = ax² + bx + c with no story attached, any real number can be used for x, so the domain is all real numbers. The range is all y ≥ k if the parabola opens up, and all y ≤ k if it opens down.

A simplified real-world model

Quadratic functions can model many situations. A common one is the height of a tossed ball. In a simple model that ignores air resistance, a ball tossed straight up has height h = −16t² + vt + s feet at t seconds, where s is its starting height in feet and v is its starting speed upward in feet per second. A real ball is affected by the air, so real heights can differ from this model.

In a story, only part of the parabola makes sense. Time cannot be negative, and the ball stops when it reaches the ground. So the domain and range in the story are smaller than all real numbers. Worked example 4 shows this.

Key words

Quadratic function: A function whose rule can be written in standard form y = ax² + bx + c, where a ≠ 0.
Parabola: The smooth U-shaped curve that is the graph of a quadratic function. It opens up if a > 0 and down if a < 0.
Vertex: The turning point of a parabola. It is the lowest point if the parabola opens up and the highest point if it opens down.
Axis of symmetry: The vertical line through the vertex that splits the parabola into two mirror-image halves. Its equation is x = −b/(2a).
y-intercept: The y-value where a graph crosses the y-axis. For y = ax² + bx + c it is c, and the graph crosses at the point (0, c).
x-intercept (zero): An x-value where the graph meets the x-axis, so y = 0 there. If it is p, the graph passes through (p, 0).
Maximum and minimum value: The largest or smallest y-value of a function. For a parabola, it is the y-value of the vertex: a maximum if the parabola opens down, a minimum if it opens up.
Domain and range: The domain is the set of all x-values you can use. The range is the set of all y-values that come out.

Worked examples

Example 1. Graph y = x² − 2x − 3 by finding its direction, axis, vertex and intercepts.

Read the numbers: a = 1, b = −2, c = −3. Since a > 0, the parabola opens up, and the vertex will be its lowest point.
Axis of symmetry: x = −b/(2a) = −(−2)/(2 · 1) = 2/2 = 1. The axis is the line x = 1.
Vertex: substitute x = 1. y = 1² − 2(1) − 3 = 1 − 2 − 3 = −4. The vertex is (1, −4).
y-intercept: c = −3, so plot (0, −3). The point x = 0 is 1 unit left of the axis, so the mirror point is 1 unit right of it, at x = 2. Check: 2² − 2(2) − 3 = −3, so (2, −3) is on the graph.
x-intercepts: set y = 0. x² − 2x − 3 = 0 factors as (x − 3)(x + 1) = 0, because (x − 3)(x + 1) = x² + x − 3x − 3 = x² − 2x − 3. A product is 0 only when one of its factors is 0, so setting each factor to 0 gives x = 3 or x = −1. The graph crosses the x-axis at (3, 0) and (−1, 0). The axis x = 1 is halfway between −1 and 3, as symmetry requires.
More points in mirror pairs: x = −2 gives y = 4 + 4 − 3 = 5, and its mirror x = 4 gives y = 16 − 8 − 3 = 5. The table for x = −2, −1, 0, 1, 2, 3, 4 has y = 5, 0, −3, −4, −3, 0, 5.
Draw one smooth curve through these points, rounded at the vertex, with arrowheads at both ends.

Answer: Answer: A parabola that opens up, with axis x = 1, vertex (1, −4), y-intercept −3 (the point (0, −3)), and x-intercepts −1 and 3.

Example 2. Graph y = −2x² + 6x − 1. The vertex is not at a whole-number x-value, so be careful.

Read a = −2, b = 6, c = −1. Since a < 0, the parabola opens down and the vertex is its highest point. Since |a| = 2 is bigger than 1, it is narrower than y = −x².
Axis: x = −6/(2 · (−2)) = −6/(−4) = 3/2. The axis is the line x = 3/2, which is 1.5.
Vertex y-value: y = −2(3/2)² + 6(3/2) − 1 = −2(9/4) + 9 − 1 = −9/2 + 8 = 7/2. The vertex is (3/2, 7/2), which is (1.5, 3.5).
y-intercept: c = −1, so plot (0, −1). It is 3/2 left of the axis, so its mirror point is 3/2 right of the axis, at x = 3. Check: −2(9) + 6(3) − 1 = −1.
Table with pairs equally far from the axis x = 3/2: x = 1 and x = 2 (each 1/2 away) both give y = 3. For x = 1: −2 + 6 − 1 = 3. And x = −1 and x = 4 (each 5/2 away) both give y = −9.
x-intercepts: the parabola opens down and its vertex is above the x-axis (7/2 is positive), so it crosses the x-axis twice. The y-values are −1 at x = 0 and 3 at x = 1, so the graph goes from below the axis to above it between x = 0 and x = 1. It comes back down between x = 2 (y is 3) and x = 3 (y is −1). The x-intercepts are not whole numbers, and this trinomial does not factor with integers, so their exact values wait for a later lesson.

Answer: Answer: A narrow parabola that opens down, with axis x = 3/2, vertex (3/2, 7/2), y-intercept −1, and two x-intercepts: one between 0 and 1 and one between 2 and 3.

Example 3. Graph y = (x + 1)(x − 5) by starting from its zeros.

Set each factor to 0: x + 1 = 0 gives x = −1, and x − 5 = 0 gives x = 5. The graph crosses the x-axis at (−1, 0) and (5, 0). (A product is 0 only when one of its factors is 0, so you can set each factor to 0. The lesson Solving Quadratics by Factoring explains why.)
The axis is halfway between the zeros: (−1 + 5)/2 = 2. So the axis is x = 2.
Vertex: y = (2 + 1)(2 − 5) = 3 · (−3) = −9. The vertex is (2, −9).
Multiply out to read a, b and c: (x + 1)(x − 5) = x² − 5x + x − 5 = x² − 4x − 5. So a = 1 (opens up), b = −4 and c = −5. Check the axis with the formula: −b/(2a) = 4/2 = 2. It matches.
y-intercept: c = −5, so plot (0, −5). Its mirror point is (4, −5).
The table for x = −1, 0, 1, 2, 3, 4, 5 has y = 0, −5, −8, −9, −8, −5, 0. For example, x = 1 gives (2)(−4) = −8. Draw a smooth curve through them.

Answer: Answer: A parabola that opens up, with x-intercepts −1 and 5, vertex (2, −9), and y-intercept −5.

Example 4. In a simplified model, a ball is tossed upward from the edge of a balcony 32 feet above the ground. It rises, then falls past the balcony to the ground below. Its height is h = −16t² + 16t + 32 feet at t seconds after the toss. Find the highest point, when the ball lands, and the domain and range that fit the story.

Read a = −16, b = 16, c = 32. Since a < 0, the graph opens down and the vertex is the highest point.
Axis: t = −16/(2 · (−16)) = −16/(−32) = 1/2. Highest height: h = −16(1/2)² + 16(1/2) + 32 = −4 + 8 + 32 = 36. In this model the ball reaches its highest point, 36 feet, half a second after the toss.
y-intercept: c = 32, so the graph crosses the vertical axis at (0, 32). At t = 0 the ball is at the balcony height, 32 feet.
Landing: the ball is on the ground when h = 0. −16t² + 16t + 32 = 0. Divide both sides by −16, which is not 0, so the solutions do not change: t² − t − 2 = 0. Factor: (t − 2)(t + 1) = 0, so t = 2 or t = −1.
Check both: h(2) = −64 + 32 + 32 = 0 and h(−1) = −16 − 16 + 32 = 0. Both are zeros of the function, and the axis t = 1/2 is halfway between −1 and 2, as it should be.
In the story, t = −1 is 1 second before the toss, and the model describes only the time after the toss. So the ball lands at t = 2 seconds, and the flight lasts from t = 0 to t = 2.
Domain in the story: 0 ≤ t ≤ 2. Range in the story: the height starts at 32, rises to 36, and falls to 0, so 0 ≤ h ≤ 36. A mirror check: the ball is back at 32 feet at t = 1, since −16 + 16 + 32 = 32.

Answer: Answer: Highest point 36 feet at t = 1/2 second. It lands at t = 2 seconds. In this model the domain for the flight is 0 ≤ t ≤ 2 and the range is 0 ≤ h ≤ 36.

Example 5. Graph y = x² + 2x + 4 and say how many x-intercepts it has.

Read a = 1, b = 2, c = 4. Since a > 0, the parabola opens up.
Axis: x = −2/(2 · 1) = −1. The sign flips: b is positive, so the axis is to the left of the y-axis.
Vertex: y = (−1)² + 2(−1) + 4 = 1 − 2 + 4 = 3. The vertex is (−1, 3).
The vertex is the lowest point, and its y-value 3 is above the x-axis. Every other point is higher, so no point has y = 0. The graph has no x-intercepts.
y-intercept: c = 4, so plot (0, 4). Its mirror point is (−2, 4). More points: x = −3 gives 9 − 6 + 4 = 7 and x = 1 gives 1 + 2 + 4 = 7.
The minimum value is 3, at x = −1. The domain is all real numbers and the range is y ≥ 3.

Answer: Answer: A parabola that opens up with vertex (−1, 3) and y-intercept 4. It has no x-intercepts. The minimum value is 3 and the range is y ≥ 3.

Example 6. Describe y = −(1/2)x² + 8 without building a long table. Find its direction, width, vertex, intercepts and range.

Read a = −1/2, b = 0, c = 8. Since a < 0, it opens down. Since |a| = 1/2 is smaller than 1, it is wider than y = −x².
Axis: x = −0/(2 · (−1/2)) = 0/(−1) = 0. With b = 0, the axis is the y-axis itself.
Vertex: at x = 0, y = 8. The vertex is (0, 8). It is also the y-intercept point (0, c), because the axis passes through x = 0.
x-intercepts: set y = 0. −(1/2)x² + 8 = 0 gives (1/2)x² = 8, so x² = 16. Two numbers have a square of 16, so x = 4 or x = −4. The graph crosses at (4, 0) and (−4, 0).
Table: x = −4, −2, 0, 2, 4 gives y = 0, 6, 8, 6, 0. For example, x = 2 gives −(1/2)(4) + 8 = 6.
The maximum value is 8, at x = 0. The domain is all real numbers, and the range is y ≤ 8.

Answer: Answer: A wide parabola that opens down, with vertex (0, 8), x-intercepts −4 and 4, maximum value 8 and range y ≤ 8.

In plain English

A quadratic function has an x² term in it, like y = x² − 4x + 3. Its graph is a smooth U-shaped curve called a parabola.

A positive number in front of x² makes the U open up. A negative number makes it open down, like a frown. A bigger number, ignoring its sign, makes the curve skinnier.

Every parabola has a middle line, called the axis of symmetry. The two halves are mirror images. The turning point on that line is the vertex. Its x-value is −b/(2a): flip the sign of b, then divide by twice a.

Where the graph crosses the y-axis is easy. It is the number c. Where it crosses the x-axis is trickier. It may cross twice, touch once, or miss the axis completely.

To draw one, find the axis and the vertex first. Then pick x-values on both sides of the axis. Points that are the same distance from the axis have the same height, so you only do half the work. Finish with one smooth curve.

The vertex also gives the highest or lowest value. That is why a parabola can describe the top of a ball’s flight or the best price at a lemonade stand.

See it

For y = x² − 2x − 3: the axis is x = 1, the vertex is (1, −4), the y-intercept point (0, −3) has the mirror point (2, −3), and the x-intercepts are −1 and 3.

The sign of a sets the direction and the size of |a| sets the width. At x = 1 the four heights are 2, 1, 1/2 and −1.

The same idea, the fun way

Pick the one that clicks. Each one explains the same concept from a different angle.

Sports

The basketball arc

A player lofts a basketball toward the hoop. It climbs, slows its climb, reaches its highest point, and then drops. Look at the height only. In a simplified model with no air and no spin, the ball’s height a short time before the top equals its height the same short time after the top. For example, 0.3 seconds before the peak and 0.3 seconds after it, the ball is at the same height.

How it connects: The height model is a quadratic in the time t with a < 0, so its graph opens down. The top of the arc is the vertex, and the maximum height happens at t = −b/(2a). The matching heights before and after the top are mirror points on either side of the axis of symmetry. Limit: this is a model. Air and spin change a real shot. Also, the parabola is the graph of height against time, so its horizontal axis shows time, not distance across the court.

Gaming

The platformer jump

In a made-up platform game, the hero’s height above the ground, in blocks, is h = −4t² + 8t at t seconds after the jump starts. At t = 0 the height is 0. At t = 1 it is 4 blocks, and at t = 2 the hero is back on the ground. At t = 1/2 and at t = 3/2, the height is 3 blocks both times.

How it connects: Since a = −4 is negative, the jump arc opens down. The axis of symmetry is t = −8/(2 · (−4)) = 1, so the vertex (1, 4) is the top of the jump. Take-off (t = 0) and landing (t = 2) are the zeros, and they are mirror images across the axis. The times t = 1/2 and t = 3/2 are each 1/2 away from the axis, so the heights match. Limit: the game is invented, and the formula describes only the time in the air. The parabola keeps going on both sides, but the hero is not there.

Relationships

Meet halfway

Two friends live along one straight road, one at mile marker 4 and one at mile marker 12. They agree to meet at the point exactly halfway between their homes, at mile marker 8. Each of them walks 4 miles.

How it connects: When a parabola crosses the x-axis at two points, its axis of symmetry runs exactly halfway between them. For y = (x − 4)(x − 12), the x-intercepts are 4 and 12, the axis is x = (4 + 12)/2 = 8, and the vertex is (8, −16). Limit: this shortcut needs two x-intercepts. A parabola with none still has an axis, and then you use −b/(2a). Also, halfway tells you only the vertex’s x-coordinate. You still substitute to find its y-coordinate.

Food

The lemonade stand

In a made-up rule, a stand that charges p dollars per cup sells 30 − 5p cups. Revenue is the money collected before any costs: R = p(30 − 5p) = −5p² + 30p dollars. At 1 dollar per cup the stand sells 25 cups and collects 25 dollars. At 2 dollars it sells 20 cups and collects 40. At 3 dollars: 15 cups, 45 dollars. At 4 dollars: 10 cups, 40 dollars. At 5 dollars: 5 cups, 25 dollars. At 6 dollars: 0 cups, 0 dollars.

How it connects: The table is symmetric: 2 and 4 dollars give the same revenue, and so do 1 and 5. The axis is p = −30/(2 · (−5)) = 3, and the vertex (3, 45) says that the best price in this model is 3 dollars, with a maximum revenue of 45 dollars. The zeros p = 0 and p = 6 are the prices that bring in no money. Limit: the rule is invented, and real customers are not that tidy. Revenue is not profit, because costs are left out. And only prices from 0 to 6 dollars make sense here, so the domain in the story is smaller than all real numbers.

Proverb

What goes up must come down

The proverb says that whatever rises will eventually fall back.

How it connects: A parabola that opens down matches this: it rises to the vertex and then falls, and the vertex is the top. The proverb is only a general saying, though. It does not say how high or when, and it does not say that the fall mirrors the climb. The parabola does: in the simplified ball model, the ball is at the same height at times equally far before and after the top, and the vertex tells you exactly when the top happens and how high it is. Limit: the saying does not fit every graph. After its vertex, a parabola that opens up keeps rising forever.

Everyday joke

Looking up

Why was the upward-opening parabola so optimistic at its vertex? From there on, things were only looking up.

How it connects: Reading left to right, a parabola that opens up goes down until the vertex and then goes up, so its vertex is the lowest point and the y-values only increase after it. Limit: the pun works only for a parabola that opens up. A parabola that opens down reaches its vertex at the top, and after it the y-values only decrease.

Memory help

Acronym

AVOID

AVOID guessing when you graph y = ax² + bx + c. Axis: x = −b/(2a). Vertex: put that x back in to get y. Opens: up if a > 0, down if a < 0. Intercepts: plot (0, c) and its mirror point, and find the x-intercepts if there are any. Draw one smooth curve through the points.

Rhyme

Over two a

“Over two a goes the opposite of b: that’s the x where the vertex will be.” The opposite of b means change its sign: if b = −6 use 6, and if b = 6 use −6. Then divide by the whole of 2a.

Picture it

Smile or frown

Picture a face. When a is positive the parabola smiles (opens up), and when a is negative it frowns (opens down). Only the sign of a draws the face. A positive b or c does not make it smile.

Sketch it

Fold the paper

Sketch a parabola on paper and fold the paper so the two halves land on top of each other. The crease is the axis of symmetry, and the crease touches the curve at the vertex. Every point on one half lands on a matching point on the other half, at the same height.

Trap alert

The sign flip on the axis

For y = x² + 6x + 2, b = 6 and the axis is x = −6/2 = −3, not 3. When a and b have the same sign, the axis is to the left of the y-axis (negative x). When they have opposite signs, the axis is to the right of it. Check the sign before you plot.

Trap alert

A vertex needs two numbers

The vertex is a point, written (h, k). The maximum or minimum is only the y-value, k. If the vertex is (5, 7) and the parabola opens down, the maximum value is 7, and it occurs at x = 5.

Study notes

Key ideas

A quadratic function can be written y = ax² + bx + c with a ≠ 0. Its graph is a parabola.
If a > 0 the parabola opens up and its vertex is the lowest point. If a < 0 it opens down and its vertex is the highest point. A larger |a| makes it narrower.
The axis of symmetry is the vertical line x = −b/(2a). The vertex is on it: its x-coordinate is −b/(2a), and its y-coordinate comes from substituting that x.
Points the same distance from the axis, on opposite sides, have the same y-value. Use this to plot one side and mirror it.
The y-intercept is c (the graph crosses the y-axis at (0, c)). The x-intercepts are the x-values where y = 0, and there can be two, one or none.
With no story attached, the domain is all real numbers. The range is y ≥ k if the parabola opens up and y ≤ k if it opens down, where k is the y-value of the vertex.

Steps

Put the equation in the form y = ax² + bx + c and read a, b and c. The sign of a tells you which way it opens.
Find the axis of symmetry, x = −b/(2a).
Find the vertex: substitute that x into the equation to get y.
Plot (0, c) and its mirror point across the axis. Find the x-intercepts, if there are any, by setting y = 0.
Add more points in mirror pairs, equally far from the axis.
Draw one smooth curve with arrowheads at both ends, and label the vertex and the axis.

Formulas

Standard form: y = ax² + bx + c, with a ≠ 0
Axis of symmetry: x = −b/(2a)
Vertex: (h, k) with h = −b/(2a) and k = the y-value at x = h
y-intercept: c (the graph crosses the y-axis at (0, c))
Mirror point: If the axis is x = h, the points at x and at 2h − x have the same y-value
x-intercepts: Set y = 0 and solve: two, one or no values
Range: y ≥ k if a > 0, and y ≤ k if a < 0

Watch out

Flip the sign of b in the axis formula: for y = x² + 6x + 2 the axis is x = −3.
Divide by 2a, the whole thing. For y = 2x² − 16x + 35 the axis is 16/4 = 4, not 16/2 = 8.
Only a decides whether the parabola opens up or down. The sign of b or c does not.
The maximum or minimum value is a y-value. The vertex is a point with two coordinates.
A parabola can have two, one or no x-intercepts. Do not assume it always crosses the x-axis twice.
Draw a smooth, rounded curve. Do not join the points with straight segments or draw a sharp V.
The mirror trick works only for x-values that are equally far from the axis.
In a story, use only the part of the parabola that fits. A time cannot be negative.

Why it works

Why does the sign of a decide the direction? The square x² is never negative, so ax² has the sign of a. Far from the vertex, where |x| is large, the ax² term is much bigger than bx + c, so the graph heads the way a points.
Why is the axis x = −b/(2a)? Let h = −b/(2a), so 2ah + b = 0. Compare the heights at h + d and h − d, two x-values equally far from h. Subtract the two heights. The a part is a((h + d)² − (h − d)²) = 4ahd and the b part is b((h + d) − (h − d)) = 2bd. Together, y(h + d) − y(h − d) = 4ahd + 2bd = 2d(2ah + b) = 0. So the heights match for every d, which means the graph is symmetric about the line x = h.
A quick check of the formula: y = c at x = 0 and also at x = −b/a, because ax² + bx = x(ax + b) = 0 there. The axis is halfway between these two x-values: (0 + (−b/a))/2 = −b/(2a). (If b = 0 the two x-values are the same, the axis is x = 0, and the formula agrees.)
Why can there be zero, one or two x-intercepts? A parabola changes direction only once, at the vertex. On each side of the vertex, y moves in one direction only, so each side can cross the x-axis at most once. That gives at most two x-intercepts. If the parabola opens away from the x-axis, no point reaches it (none). If the vertex is on the axis, only the vertex touches it (one). If the parabola opens toward the axis, it crosses once on each side of the vertex (two).

Common mistakes

✗ MistakeThe sign of b decides whether a parabola opens up or down.

✓ InsteadOnly the sign of a decides. If a > 0 it opens up, and if a < 0 it opens down, whatever b and c are.

✗ MistakeThe axis of symmetry of y = x² + 22x + 9 is x = 11.

✓ InsteadThe axis is x = −b/(2a) = −22/2 = −11. Flip the sign of b, and divide by twice a.

✗ MistakeThe vertex is always the y-intercept.

✓ InsteadThe y-intercept point is (0, c). The vertex is on the y-axis only when the axis is x = 0, which happens when b = 0.

✗ MistakeEvery parabola crosses the x-axis twice.

✓ InsteadA parabola can cross twice, touch once at its vertex, or never reach the x-axis. The vertex and the direction tell you which.

✗ MistakeTo graph a parabola, plot a few points and join them with straight segments.

✓ InsteadA parabola is a smooth curve. Use the vertex and mirror pairs of points, then draw one rounded curve with arrowheads, not a V or a chain of segments.

✗ MistakeThe maximum value of a parabola is the x-coordinate of the vertex.

✓ InsteadThe maximum or minimum value is the y-coordinate of the vertex. The x-coordinate tells you where it occurs.

Teach it back

Explain to a friend how to graph y = −x² + 6x − 5. Use the direction, the axis and vertex, the y-intercept and its mirror point, and the x-intercepts.

Explaining it out loud shows what you really know. Did you…

I said that a = −1 is negative, so the parabola opens down and its vertex is the highest point.
I found the axis x = −6/(2 · (−1)) = 3 and the vertex (3, 4), because −9 + 18 − 5 = 4.
I plotted the y-intercept point (0, −5) and its mirror point (6, −5).
I found the x-intercepts by setting y = 0, multiplying both sides by −1 to get x² − 6x + 5 = 0, and factoring: (x − 1)(x − 5) = 0 gives x = 1 and x = 5, each 2 units from the axis.
I drew one smooth curve with arrowheads, and said that the maximum value is 4 and the range is y ≤ 4.

Your 3-week memory plan

TodayRead the lesson, then try the flashcards.
TomorrowRecall it from memory first, then check.
In 3 daysTake the true/false and practice questions.
In 1 weekMix it with older topics and teach it back.
In 3 weeksA last quick review before it fades.

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