Graph a quadratic function and identify the vertex, axis of symmetry, intercepts, and direction of opening (LA A1: F-IF.C.7, F-IF.B.4).

What this topic is asking

Standard A1: F-IF.C.7 (with F-IF.B.4) asks you to graph a quadratic and identify its key features: the vertex, the axis of symmetry, the intercepts, and the direction of opening. On LEAP 2025 these are Type I Major Content items, including graphing items where you plot a parabola or its vertex. The axis-of-symmetry and vertex-form tools are not on the reference sheet.

Direction of opening

The leading coefficient decides which way the parabola opens:

• $a > 0$: opens upward, a U shape, vertex is the lowest point (a minimum).
• $a < 0$: opens downward, an upside-down U, vertex is the highest point (a maximum).

The larger $|a|$, the narrower the parabola.

Axis of symmetry and vertex

The axis of symmetry is the vertical line through the vertex. Find it with $x = \frac{-b}{2a}$, then get the vertex's $y$-value by substituting.

The intercepts

• $y$-intercept: set $x = 0$; it is the constant $c$. For $y = x^2 + 4x - 5$, the $y$-intercept is $-5$.
• $x$-intercepts (zeros): set $y = 0$ and solve by factoring, square roots, or the formula. For $y = x^2 + 4x - 5 = (x + 5)(x - 1)$, the $x$-intercepts are $x = -5$ and $x = 1$.

The vertex's $x$-coordinate is the midpoint of the two $x$-intercepts, a useful check.

How LEAP examines this topic

• Graphing item. Plot the parabola or place the vertex on a grid.
• Equation response. Find the axis of symmetry, vertex, or intercepts.
• Multiple choice. Identify the direction of opening, or read a feature from a graph.

A clarifying idea: vertex form $y = a(x - h)^2 + k$ shows the vertex $(h, k)$ directly, and you obtain it by completing the square, which links this topic to that method.

Why the parabola is symmetric about x = -b/2a

A parabola is symmetric about the vertical line $x = \frac{-b}{2a}$ because the quadratic produces matching outputs for inputs equally far on either side of that line, which is the structural fact behind F-IF.C.7. The two $x$-intercepts (when they exist) are given by the quadratic formula as $\frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$: notice both sit the same distance $\frac{\sqrt{b^2 - 4ac}}{2a}$ from $\frac{-b}{2a}$, one to the left and one to the right. Their midpoint is therefore exactly $\frac{-b}{2a}$, the axis of symmetry, and the vertex lies on it. This is why the axis formula works even when the parabola has no real $x$-intercepts: the center of symmetry is determined by $a$ and $b$ alone, not by where (or whether) the curve crosses the axis. It also explains a fast graphing strategy: find the axis, find the vertex on it, then use symmetry to mirror any plotted point to the other side, so one point and the vertex give you three. Recognizing the symmetry connects the algebra (the $\pm$ in the formula) to the geometry (the mirror-image curve), which is the heart of understanding a parabola.

Try this

Q1. For $y = x^2 - 2x - 3$, find the axis of symmetry. [2 points]

• Cue. $x = \frac{-(-2)}{2(1)} = 1$.

Q2. Which way does $y = 3x^2 - 1$ open, and is the vertex a max or min? [1 point]

• Cue. $a = 3 > 0$: opens up, vertex is a minimum.