Parabolas

The graph of a quadratic is a curve

A quadratic is an equation whose highest power of $x$ is a square, written in the standard form

The restriction $a \ne 0$ is what makes it quadratic: if $a$ were $0$ the $x^2$ term would disappear and you would be left with the line $y = bx + c$. With a genuine $x^2$ term the graph is no longer straight. To find out what it is, do exactly what you did for a line: build a table of solutions, plot the points, and connect them.

Start with the plainest quadratic there is, $y = x^2$, where $a = 1$, $b = 0$, and $c = 0$. Pick a spread of $x$-values and square each one.

Plot the five points. They do not line up; they bend. Joining them with a smooth curve instead of straight segments gives a rounded U. That U dips to a lowest point at the origin and sweeps upward on both sides. That curve is a parabola, and every quadratic graphs as one.

Look again at the table, because it hides the parabola’s most important secret. The rows pair up: $x = -2$ and $x = 2$ both give $y = 4$, and $x = -1$ and $x = 1$ both give $y = 1$. Squaring throws away the sign, so two inputs that are opposite in sign land at exactly the same height. That is why the left half of the curve is a mirror image of the right half, and it is the first fact we will make precise below.

The same table-and-plot routine graphs any quadratic. When $b$ or $c$ is not zero the bowl slides around the plane, but its shape never changes.

Which way it opens, and how wide

The single number $a$, the coefficient of $x^2$, controls both the direction the parabola opens and how wide it is. You can read the reason straight off $y = x^2$.

Every output of $y = x^2$ is a square, so $y$ is never negative, and $y = 0$ happens only at $x = 0$. As $x$ moves away from $0$ in either direction, $x^2$ grows, so the curve climbs on both sides. That is why it forms a valley with its lowest point at the origin. Now bring back the coefficient $a$. Multiplying by a positive $a$ only rescales those heights, so a valley stays a valley and the parabola opens upward. Multiplying by a negative $a$ flips the sign of every height, turning the valley upside down into a hill, so the parabola opens downward.

The size of $a$ sets the steepness. A large $\lvert a \rvert$ multiplies each height by more, so the arms shoot up faster and the parabola looks narrow. A small $\lvert a \rvert$ scales each height down, so the arms rise gently and the parabola looks wide.

The axis of symmetry

That mirror pattern in the table is worth pinning down, because it is the key to everything else. A parabola is symmetric about a vertical line, its axis of symmetry, and for $y = ax^2 + bx + c$ that line is

For $y = x^2$, where $b = 0$, the axis is $x = 0$, the y-axis itself, which matches the fold you saw in the table. The reason the axis sits at $-\tfrac{b}{2a}$ comes straight from completing the square, the tool you learned two lessons ago.

The vertex

The axis of symmetry passes through one special point on the parabola, the place where the curve stops falling and starts rising (or stops rising and starts falling). That turning point is the vertex, and it sits right where the axis meets the curve, at

To get the matching $y$, substitute that $x$ back into the equation. There is nothing new to memorize for the y-coordinate: find the x-coordinate from $-\tfrac{b}{2a}$, then plug it in.

Because the axis is the fold line, the vertex is the one point that has no mirror twin, so it is the turning point of the whole curve. When the parabola opens upward ($a > 0$) the vertex is the lowest point on the graph, and when it opens downward ($a < 0$) the vertex is the highest point. You can see why from the completed-square form $y = a\left(x + \tfrac{b}{2a}\right)^2 + \left(c - \tfrac{b^2}{4a}\right)$, where the square is never negative. So for $a > 0$ the term $a(\ldots)^2$ only adds to the constant, making $y$ smallest when the square is $0$, which is exactly at the vertex. For $a < 0$ the same term only subtracts, so $y$ is largest there. (Reading a real-world lowest cost or highest point off a parabola is the job of a later lesson; here it is simply a fact about the shape.)

Vertex form

The completed-square form from the proof above is so useful for graphing that it has its own name. Vertex form is

and its whole point is that you can read the vertex straight off it: the vertex is $(h, k)$. Set $x = h$ and the square is $0$, so $y = k$, which is the turning value. The leading coefficient $a$ is also the same $a$ as in standard form, so it still tells you the direction and width. Matching this against the completed square shows $h = -\tfrac{b}{2a}$ and $k = c - \tfrac{b^2}{4a}$, the same axis and vertex height as before.

Watch the sign inside the parentheses. Vertex form is $x - h$, so $y = (x - 3)^2 + 2$ has $h = 3$ (vertex x-coordinate $3$), while $y = (x + 3)^2 + 2$ is really $(x - (-3))^2 + 2$, so $h = -3$. The number inside the parentheses is the opposite of the vertex’s x-coordinate.

Converting standard form to vertex form is just completing the square; converting back is just expanding. Both directions are worth practicing, because each form shows you something the other hides. Standard form hands you $c$ as the y-intercept, and vertex form hands you the vertex.

The y-intercept

The y-intercept is where the graph crosses the y-axis, and every point on the y-axis has $x = 0$. Set $x = 0$ in $y = ax^2 + bx + c$ and the first two terms vanish:

So the y-intercept is the point $(0, c)$, read off instantly as the constant term of standard form. For $y = x^2 - 2x - 3$ the y-intercept is $(0, -3)$; for $y = -x^2 + 4$ it is $(0, 4)$. Every parabola has exactly one y-intercept, because a vertical line meets the curve just once.

The x-intercepts and the discriminant

The x-intercepts are where the graph crosses the x-axis, and every point on the x-axis has $y = 0$. So to find them, set $y = 0$ and solve the quadratic equation

by factoring when it splits nicely, or by the quadratic formula when it does not. Each real solution is one x-intercept. Unlike the y-intercept, a parabola can have two, one, or zero x-intercepts. Which of the three you get is decided entirely by the discriminant $b^2 - 4ac$ that you met with the quadratic formula.

That argument is worth watching happen. The graph below is $y = ax^2 + bx + c$ with all three coefficients under your control, and it marks what you would otherwise have to compute. Those marks are the vertex, the dashed axis of symmetry through it, and a ring at each crossing of the horizontal axis.

Start from $c = -3$ and step $c$ upward. The discriminant here is $4 - 4c$, so it falls as $c$ rises, and the two rings slide toward each other. At $c = 1$ they meet: the discriminant is exactly $0$, the two crossings have become one, and it is sitting on the vertex. One more step and both rings are gone, because the whole curve has lifted off the axis. Nothing about the parabola’s shape changed on the way; only its height did, which is the argument above run backwards.

Two more things to try. Move $b$ and watch the axis of symmetry slide while the shape stays identical, which is $-\tfrac{b}{2a}$ doing its work. And take $a$ to $0$: the squared term disappears and what is left is a straight line. That is the plainest possible answer to why the definition of a quadratic insists $a \ne 0$.

Putting it all together

Each feature is quick on its own; together they let you sketch a parabola without a long table. The recipe is always the same: read the direction from the sign of $a$, then find the vertex from $-\tfrac{b}{2a}$. Mark the y-intercept at $(0, c)$, and find the x-intercepts by setting $y = 0$.

Common mistakes