Quadratics, Parabolas, and the Discriminant

A quadratic is any function whose highest power of the variable is two, and its graph is always a parabola: a single smooth curve with one turning point and perfect left-right symmetry about a vertical line through that turning point. Every quadratic question you will see is a question about one of four features, namely the direction it opens, where it turns, where it crosses the x-axis, and where it crosses the y-axis.

The SAT tests quadratics constantly in the Advanced Math domain, usually by writing the same function in a form that hides the feature you need and asking you to reveal it. A question about the minimum value of a profit function will hand you standard form. A question about the x-intercepts will hand you vertex form. Recognizing that the work is a form conversion, not a new technique, is most of the battle. The ACT covers the same ground and adds more direct algebra: factoring, the quadratic formula, and the number of real solutions of an equation.

The most common way students lose points here is by memorizing three unconnected procedures. Standard form, vertex form, and factored form are the same function; they differ only in what they display for free. Once you can say what each form gives away instantly, you stop guessing which method to start with, and questions that look unfamiliar become routine.

A quadratic function in standard form is y = ax² + bx + c with a not equal to zero. The sign of a decides which way the parabola opens: when a > 0 the curve opens upward and the turning point is a minimum, and when a < 0 it opens downward and the turning point is a maximum. The size of a controls how narrow the curve is, because multiplying the squared term by a larger number makes the outputs climb faster for the same input. The constant c is the y-intercept, since substituting x = 0 wipes out the first two terms and leaves y = c.

Every parabola is symmetric about a vertical line called the axis of symmetry, and in standard form that line is $x = -\frac{b}{2a}$. The reason is visible in the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: the two solutions sit the same distance, $\frac{\sqrt{b^2 - 4ac}}{2a}$, on either side of the number -b / (2a). That center point is the x-coordinate of the vertex whether or not the parabola actually crosses the axis. To get the y-coordinate of the vertex, substitute that x back into the function. This is the fastest route to a maximum or minimum in any word problem that hands you standard form.

Vertex form is y = a(x - h)² + k, and it displays the vertex (h, k) directly. It works because (x - h)² is a square, so it is never negative and equals zero only when x = h. If a is positive, the smallest possible value of a(x - h)² is zero, so the smallest possible y is k, achieved exactly at x = h. If a is negative, the same reasoning runs the other way and k is the maximum. Watch the sign convention: y = (x + 5)² - 18 has vertex (-5, -18), because x + 5 is x - (-5).

Factored form is y = a(x - r)(x - s), and it displays the x-intercepts, r and s, because a product equals zero exactly when one of its factors equals zero. This zero product property is the reason factoring solves equations at all. Factored form also gives the vertex almost for free: by symmetry the axis must sit exactly halfway between the two roots, so $h = \frac{r + s}{2}$. That shortcut is worth memorizing, because reading roots off a factored expression and averaging them is much faster than expanding and using -b / (2a).

Completing the square is the bridge from standard form to vertex form. Given y = x² + bx + c, take half of b, square it, and add and subtract that number so nothing changes: $y = \left(x^2 + bx + \left(\tfrac{b}{2}\right)^2\right) - \left(\tfrac{b}{2}\right)^2 + c$, and the parenthesized part is exactly $\left(x + \tfrac{b}{2}\right)^2$. The reason half of b is the magic number is that expanding (x + p)² produces x² + 2px + p², so matching the linear terms forces 2p = b, that is p = b / 2. When the leading coefficient is not 1, factor it out of the first two terms first, and remember that anything you add inside those parentheses gets multiplied by that factor when you compensate outside.

The discriminant is the expression under the radical in the quadratic formula, D = b² - 4ac, and it counts real solutions without requiring you to find them. If D > 0 the square root is a positive real number and there are two distinct real solutions, so the parabola crosses the x-axis twice. If D = 0 the plus-or-minus adds nothing and there is exactly one real solution, a repeated root, so the parabola touches the axis at its vertex. If D < 0 there is no real square root and there are no real solutions, so the parabola misses the x-axis entirely. Question types to recognize: any prompt containing the phrases exactly one solution, no real solutions, or the graph is tangent to the x-axis is a discriminant question in disguise, and any prompt asking for a greatest value, least value, maximum height, or maximum revenue is a vertex question.

In the interactive graph below, drag a to see the parabola open upward or downward and get narrower or wider, and notice that the y-intercept never moves while you do it. Drag c and watch the whole curve slide vertically until it stops crossing the x-axis, which is the moment the discriminant readout turns negative and the two x-intercept markers disappear. Drag b and track how the vertex marker slides along, since its x-coordinate is always -b / (2a).

Reading the vertex with the wrong sign from vertex form. In y = a(x - h)² + k the vertex is (h, k), so y = 2(x + 3)² - 5 has vertex (-3, -5), not (3, -5). Rewrite the plus as a minus of a negative in your head before you read off h, and confirm by checking that your h makes the squared term equal zero.

Forgetting the 2a in the axis of symmetry. The axis is x = -b / (2a), not x = -b / a. When a is 1 the two formulas agree, which is exactly why the error survives practice and then costs points on a problem where a is 2 or -16. Write the formula out with the parentheses every time.

Stopping at the x-coordinate of the vertex. A question asking for the maximum height, the minimum cost, or the greatest value wants the output, k, but -b / (2a) gives only the input, h. You must substitute back. If a question asks when the maximum occurs, it wants h; if it asks what the maximum is, it wants k. Reread the last line of the prompt before choosing.

Misreading the discriminant for a tangent graph. When a problem says a parabola is tangent to the x-axis, or touches it at exactly one point, the discriminant is zero, not positive. Students often set b² - 4ac > 0 out of habit. Translate the phrase into a count of real solutions first, then pick the sign.

Dropping the leading coefficient while completing the square. In y = 3x² + 12x + 1, factoring gives y = 3(x² + 4x) + 1, and adding 4 inside the parentheses actually adds 3 times 4, so you must subtract 12 outside, giving y = 3(x + 2)² - 11. Compensating with 4 instead of 12 is the single most common completing-the-square error; always expand your final answer to check it matches the original.