Quadratic functions and their graphs: use standard, factored and vertex form to read the y-intercept, the x-intercepts and the vertex, and connect the discriminant to the number of x-intercepts.

What this skill is asking

A quadratic function $f(x) = ax^2 + bx + c$ graphs as a parabola. The Digital SAT (Advanced Math domain) tests reading a parabola's key features (vertex, axis of symmetry, $x$-intercepts, $y$-intercept) and moving between the three forms of a quadratic, each of which displays a different feature directly. It also connects the discriminant to how many times the parabola crosses the $x$-axis.

The three forms and what they show

Each form is the right tool for a different question.

A worked feature read

Converting to the right form exposes the answer.

The discriminant and x-intercepts

The number of times a parabola crosses the $x$-axis is the number of real zeros, which the discriminant $b^2 - 4ac$ decides. A positive discriminant means two $x$-intercepts (the parabola crosses twice). A zero discriminant means one $x$-intercept (the parabola touches the axis at its vertex, a repeated root). A negative discriminant means no $x$-intercepts (the parabola sits entirely above or below the axis). SAT questions often phrase this geometrically ("the graph touches the $x$-axis at exactly one point") and expect you to translate it into the discriminant condition and solve for a parameter.

Symmetry and the vertex as an optimum

A parabola is symmetric about the vertical line through its vertex, so the two $x$-intercepts are equidistant from $x = h$, and any two points with the same $y$-value share that symmetry. This makes the vertex the optimum: the minimum value of the function when the parabola opens up, or the maximum when it opens down. Real-world SAT questions use this for "maximum height", "minimum cost", or "greatest area" problems: put the quadratic in vertex form (or find $x = -\frac{b}{2a}$ and evaluate) and read the optimum value $k$ at the vertex. Desmos confirms all of this instantly by graphing and clicking the vertex or intercepts.

Finding the vertex from the intercepts

The symmetry also gives a shortcut when you know the zeros. Because the axis of symmetry sits exactly halfway between the two $x$-intercepts, the $x$-coordinate of the vertex is the average of the roots. If a parabola has zeros at $x = 2$ and $x = 8$, the axis of symmetry is $x = \frac{2 + 8}{2} = 5$, and evaluating the function at $x = 5$ gives the vertex's $y$-value. This is often faster than computing $-\frac{b}{2a}$, especially when the quadratic is already in factored form. It also explains a frequently tested fact: a quadratic's graph reaches its optimum midway between its zeros, so a projectile that leaves and returns to the ground at two times is at its highest exactly halfway between them. Recognising the vertex as the midpoint of the roots ties the algebra (zeros) to the geometry (symmetry) in one move.