Graph quadratic functions on the coordinate plane and identify key attributes, including x-intercept, y-intercept, zeros, maximum or minimum value, vertex, and the axis of symmetry (TEKS A.7A, A.3B).

What this topic is asking

Quadratics are about a quarter of the STAAR Algebra I test, and graphing them is the foundation. TEKS A.7A (with A.3B) asks you to graph a quadratic function and read its key attributes: the vertex, axis of symmetry, $x$-intercepts (zeros), $y$-intercept, and the maximum or minimum value. On the redesigned test this is assessed by traditional questions and by hot-spot graphing, where you plot the vertex or other points directly.

The parabola and its direction

A quadratic $f(x) = ax^2 + bx + c$ graphs as a parabola, a symmetric U-shaped curve. The coefficient $a$ controls the opening:

• $a > 0$: opens upward, and the vertex is the lowest point (a minimum).
• $a < 0$: opens downward, and the vertex is the highest point (a maximum).

The larger $|a|$, the narrower the parabola; the smaller $|a|$, the wider.

Vertex and axis of symmetry from standard form

In standard form $f(x) = ax^2 + bx + c$, the axis of symmetry is the reference-sheet line $x = \frac{-b}{2a}$. The vertex sits on this line: its $x$-coordinate is $\frac{-b}{2a}$, and its $y$-coordinate is found by substituting that $x$ back into the function.

Intercepts, zeros, and vertex form

The $y$-intercept is $f(0) = c$, the constant term. The zeros (or $x$-intercepts) are where $f(x) = 0$, found by factoring or the quadratic formula. The parabola is symmetric about its axis, so the two zeros are equally spaced on either side of the vertex's $x$-value.

In vertex form $f(x) = a(x - h)^2 + k$, the vertex is read directly as $(h, k)$. Watch the sign: $(x - 3)^2$ gives $h = +3$, while $(x + 3)^2 = (x - (-3))^2$ gives $h = -3$.

How STAAR examines this topic

• Multiple choice. Find the axis of symmetry or vertex, match a function to its graph, or identify the maximum or minimum. The "$-b$ sign" and "$h$ sign" errors are standard distractors.
• Hot spot. Plot the vertex or zeros on a grid; placement must be exact.
• Inline choice. Choose opens up or down, maximum or minimum, from dropdowns.

A clarifying idea is that the axis of symmetry organizes the whole graph: once you know it and the vertex, every other point comes in mirror-image pairs, so a single point to one side gives you its partner on the other side for free.

A reliable five-point graphing routine

To sketch any parabola accurately enough for a hot-spot item, build five points. First, find the axis of symmetry $x = \frac{-b}{2a}$ and the vertex by substituting that $x$ back in. Second, plot the $y$-intercept $(0, c)$. Third, use symmetry to mirror the $y$-intercept across the axis: if the axis is $x = 2$ and the $y$-intercept is at $x = 0$ (two units left of the axis), its mirror is two units right, at $x = 4$, with the same $y$-value. Those three points plus the vertex usually determine the curve, and a fifth point one step further out confirms the width.

This routine also explains why the zeros are symmetric about the axis. If a parabola has zeros at $x = -1$ and $x = 5$, the axis of symmetry is exactly halfway between them, at $x = \frac{-1 + 5}{2} = 2$, and the vertex sits on that line. Reading the vertex's $x$-coordinate as the average of the two zeros is a fast alternative to the formula whenever the zeros are known, and it is a check the test rewards on items that give a factored function or a graph with visible intercepts.

Try this

Q1. Find the axis of symmetry of $f(x) = 2x^2 + 8x - 1$. [1 point]

• Cue. $x = \frac{-8}{2(2)} = -2$.

Q2. State the vertex of $f(x) = (x + 1)^2 - 4$. [1 point]

• Cue. $h = -1$, $k = -4$, so $(-1, -4)$.