Graph quadratic functions and identify key features (vertex, axis of symmetry, zeros, y-intercept, maximum or minimum); relate the three forms; and describe the effect of transformations on the parent function.

What this topic is asking

The Regents Algebra I exam (the Interpreting Functions, F-IF, and Building Functions, F-BF, clusters) wants you to graph a quadratic, identify its vertex, axis of symmetry, zeros, y-intercept, and maximum or minimum, relate the three algebraic forms, and describe how transformations move the parent parabola $y = x^2$. A projectile-motion or area model is a staple of Part III and Part IV.

The shape and the axis of symmetry

Every quadratic $y = ax^2 + bx + c$ graphs as a parabola, symmetric about a vertical line through its turning point.

The vertex is the single most useful point: it is the maximum height of a thrown object, the minimum of a cost function, or the turning point of any parabola. Find the $x$-coordinate with the axis formula, then substitute to get the $y$-coordinate.

Intercepts and the three forms

The y-intercept is the constant $c$ in standard form, the value at $x = 0$. The zeros (x-intercepts) are where $y = 0$, found by factoring or the quadratic formula. Each algebraic form reveals a different feature:

• Standard $y = ax^2 + bx + c$ shows the y-intercept $c$.
• Factored $y = a(x - r_1)(x - r_2)$ shows the zeros $r_1$ and $r_2$.
• Vertex $y = a(x - h)^2 + k$ shows the vertex $(h, k)$.

Transformations of the parent parabola

Starting from $y = x^2$, each change to the equation moves or reshapes the graph:

• $y = x^2 + k$ shifts up by $k$ (down if $k$ is negative).
• $y = (x - h)^2$ shifts right by $h$ (left if $h$ is negative): note the sign flips.
• $y = -x^2$ reflects over the x-axis, opening downward.
• $y = ax^2$ with $|a| > 1$ makes it narrower; $0 < |a| < 1$ makes it wider.

A clarifying point worth stressing is the horizontal shift sign: $(x - 3)^2$ moves the graph 3 units to the right, even though it looks like subtraction, because the vertex is where the inside equals zero, at $x = 3$. This is the single most common transformation error, so always read $h$ as the value that makes the squared term zero.

A second useful idea is that the axis of symmetry pairs up points. Because a parabola is a mirror image across its axis, any two points with the same $y$-value are equidistant from the axis, and their $x$-coordinates average to the axis value. If a parabola passes through $(1, 0)$ and $(7, 0)$, the axis is halfway between them at $x = 4$, which is a fast way to find the vertex from the zeros without the formula. This symmetry also means that once you have plotted the vertex and one other point, you get a third point for free by reflecting across the axis, which makes a hand-drawn sketch quicker and more accurate.

Try this

Q1. What is the vertex of $y = (x + 2)^2 - 7$? [1 credit]

• Cue. Vertex form gives $(h, k) = (-2, -7)$.

Q2. Does $y = -2x^2 + 8x$ open up or down, and what is its axis of symmetry? [2 credits]

• Cue. $a = -2 < 0$ opens down; axis $x = -\frac{8}{2(-2)} = 2$.