How do I find the vertex of a parabola from standard form?

Find the axis of symmetry, x equals negative b over 2a, then substitute that x-value into the function to get the y-coordinate. The vertex is that ordered pair. The sign of a tells you the rest: if a is positive the parabola opens up and the vertex is a minimum; if a is negative it opens down and the vertex is a maximum.

What does vertex form tell me directly?

Vertex form is f of x equals a times the quantity x minus h squared, plus k, and the vertex is the point h, k. The horizontal shift is opposite the sign inside the parentheses, so x minus 4 shifts right 4. A negative a reflects the parabola downward, and the size of a stretches it narrower (greater than 1) or wider (between 0 and 1).

Why must a quadratic equal zero before I factor to solve it?

Factoring solves an equation through the zero-product property, which says that if a product equals zero then at least one factor equals zero. That logic only works when the product equals zero, so x squared plus 5x equals 24 must first become x squared plus 5x minus 24 equals 0. Factoring while one side is 24 tells you nothing, because 24 is not zero.

When do I complete the square versus use the quadratic formula?

Factor first if integer factors exist. Use the square-root property when there is no linear term or the equation is already squared. Complete the square when you are asked to, or when you also need the vertex, since it produces vertex form. Use the quadratic formula when nothing factors; it always works but involves the most arithmetic.

What does the discriminant tell me?

The discriminant is b squared minus 4ac, the part under the radical in the quadratic formula. A positive discriminant gives two real solutions and two x-intercepts, a zero discriminant gives one solution (a double root, vertex on the axis), and a negative discriminant gives no real solutions (the parabola misses the x-axis). It counts solutions without solving.

In a quadratic word problem, when do I use the vertex versus the zeros?

The vertex answers an extreme value: maximum height, maximum revenue, or minimum cost, signaled by words like how high, maximum, or least. The zeros answer a boundary event: when an object lands, hits the ground, or breaks even. Always interpret the result, rejecting negative times, lengths, or counts, and state the answer with units.

GeorgiaMaths

Georgia Milestones Algebra: a complete guide to quadratic functions and equations

A deep-dive Georgia Milestones Algebra: Concepts & Connections guide to quadratic functions and equations, spanning the Algebra and Functions domains. Covers graphing parabolas and key features, transformations and vertex form, solving by factoring, by square roots and completing the square, the quadratic formula and discriminant, and modeling with quadratics.

Generated by Claude Opus 4.817 min readA.PAR, A.FGRUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this part of the course demands
Graphing quadratic functions
Transformations and vertex form
Solving by factoring
Completing the square and square roots
The quadratic formula and the discriminant
Modeling with quadratics
How this strand is examined
Check your knowledge

What this part of the course demands

This guide covers Quadratic Functions and Equations, which spans the Algebra (A.PAR) and Functions (A.FGR) domains and is, with linear content, where the Proficient and Distinguished levels are usually decided on the EOC. The function side (graphing, transformations, modeling) and the solving side (factoring, square roots, completing the square, the formula) reinforce each other: the solutions of a quadratic equation are the x-intercepts of the parabola. Each dot-point page carries its own worked Milestones-style questions: graphing quadratic functions, transformations of quadratics, solving by factoring, completing the square and square roots, the quadratic formula and discriminant, and modeling with quadratics.

Graphing quadratic functions

A quadratic $f(x) = ax^2 + bx + c$ graphs as a parabola: $a > 0$ opens up (minimum vertex), $a < 0$ opens down (maximum vertex). The axis of symmetry is $x = \frac{-b}{2a}$ , the vertex sits on it (substitute to get $y$ ), the y-intercept is $c$ , and the x-intercepts are the zeros.

Transformations and vertex form

Vertex form $f(x) = a(x - h)^2 + k$ has vertex $(h, k)$ . The transformations from $y = x^2$ are a horizontal shift by $h$ (opposite the sign inside), a vertical shift by $k$ , a reflection if $a < 0$ , and a stretch by $|a|$ . Set the inside to zero to find $h$ .

Solving by factoring

Write the equation in standard form equal to zero, factor, and apply the zero-product property: each factor set to zero gives a solution, the opposite sign of its constant. The solutions are the x-intercepts of the parabola.

Completing the square and square roots

The square-root property: $(x - h)^2 = k$ gives $x = h \pm \sqrt{k}$ (keep the $\pm$ ). Completing the square adds $\left(\frac{b}{2}\right)^2$ to form a perfect square, solving any quadratic and converting standard form to vertex form.

The quadratic formula and the discriminant

The formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ solves every quadratic; watch the sign of $-4ac$ and simplify radicals. The discriminant $b^2 - 4ac$ counts real solutions: positive (two), zero (one), negative (none), matching the x-intercepts.

Modeling with quadratics

For projectile motion $h(t) = -16t^2 + v_0 t + h_0$ and area models, use the vertex for a maximum or minimum and the zeros for boundary events (landing, break-even). Always reject unrealistic roots (negative time, length, count) and state units.

A full modeling problem

A ball's height is $h(t) = -16t^2 + 40t + 6$ feet. Find when it lands (to the nearest tenth) and its maximum height.

step 1 Maximum height at the vertex

$t = \frac{-40}{2(-16)} = 1.25$ s; $h(1.25) = -16(1.5625) + 40(1.25) + 6 = -25 + 50 + 6 = 31$ feet.

step 2 Set height to zero for landing

$-16t^2 + 40t + 6 = 0$ , with $a = -16$ , $b = 40$ , $c = 6$ .

step 3 Apply the quadratic formula

$t = \frac{-40 \pm \sqrt{1600 + 384}}{-32} = \frac{-40 \pm \sqrt{1984}}{-32}$ , and $\sqrt{1984} \approx 44.5$ .

step 4 Choose the realistic root

$t = \frac{-40 - 44.5}{-32} \approx 2.6$ s; the negative root is rejected. Maximum height 31 feet, lands at about 2.6 seconds.

How this strand is examined

Hot spot / graphing. Plot a vertex, click x-intercepts, or match an equation to a parabola.
Numeric entry. Solve by any method in simplest radical form, or find a vertex or discriminant.
Multiple choice / inline choice. Direction of opening, number of solutions, vertex versus zeros.
Constructed response. Solve and interpret a model, or complete the square for the vertex.

Check your knowledge

Work these as you would for credit on the EOC.

For $f(x) = x^2 - 8x + 7$ , find the vertex. (2 points)
State the vertex of $g(x) = 2(x - 3)^2 - 5$ . (1 point)
Solve $x^2 - 2x - 15 = 0$ by factoring. (1 point)
Solve $(x + 1)^2 = 25$ . (1 point)
Solve $x^2 + 6x + 4 = 0$ by completing the square. (2 points)
Solve $2x^2 + 3x - 2 = 0$ using the quadratic formula. (2 points)
How many real solutions does $x^2 + x + 4 = 0$ have? (1 point)
A ball's height is $h(t) = -16t^2 + 48t$ . When does it land? (2 points)

Solutions

Step 1: Find the vertex of a parabola from standard form

The axis of symmetry formula $x = \frac{-b}{2a}$ gives the x-coordinate of the vertex; substituting that value into the function gives the y-coordinate. The vertex is the minimum when $a > 0$ and the maximum when $a < 0$ .

x = \frac{8}{2(1)} = 4, \quad f(4) = 16 - 32 + 7 = -9

Final answer: Vertex $(4, -9)$ .

Step 2: Read the vertex from vertex form

Vertex form $f(x) = a(x - h)^2 + k$ reveals the vertex $(h, k)$ directly. Note that $h$ is opposite in sign to what appears inside the parentheses.

$g(x) = 2(x - 3)^2 - 5$ has $h = 3$ and $k = -5$ .

Final answer: Vertex $(3, -5)$ .

Step 3: Solve a quadratic by factoring

Set the equation equal to zero, factor the quadratic, then apply the zero-product property: if a product equals zero, at least one factor equals zero. Each factor set to zero gives one solution.

x^2 - 2x - 15 = 0 \Rightarrow (x - 5)(x + 3) = 0

x = 5 \text{ or } x = -3

Final answer: $x = 5$ or $x = -3$ .

Step 4: Solve using the square-root property

When an equation has the form $(x - h)^2 = k$ , take the square root of both sides and include both the positive and negative roots with the $\pm$ symbol. This avoids expanding and re-factoring.

x + 1 = \pm 5 \Rightarrow x = 4 \text{ or } x = -6

Final answer: $x = 4$ or $x = -6$ .

Step 5: Solve by completing the square

Completing the square converts a quadratic into vertex form by adding $\left(\frac{b}{2}\right)^2$ to both sides, creating a perfect square trinomial on the left. This works for any quadratic and is especially useful when the quadratic formula would give a messy result.

x^2 + 6x = -4 \Rightarrow x^2 + 6x + 9 = 5 \Rightarrow (x + 3)^2 = 5 \Rightarrow x = -3 \pm \sqrt{5}

Final answer: $x = -3 \pm \sqrt{5}$ .

Step 6: Solve using the quadratic formula

The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ works for any quadratic and is the reliable fallback when factoring is not straightforward. Identify $a$ , $b$ , and $c$ from standard form before substituting.

With $a = 2$ , $b = 3$ , $c = -2$ :

x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}

x = \frac{1}{2} \text{ or } x = -2

Final answer: $x = \dfrac{1}{2}$ or $x = -2$ .

Step 7: Use the discriminant to count real solutions

The discriminant $b^2 - 4ac$ is the expression under the radical in the quadratic formula. A negative discriminant means the square root is of a negative number, which has no real value, so there are no real solutions.

b^2 - 4ac = 1 - 4(1)(4) = 1 - 16 = -15 < 0

Final answer: No real solutions.

Step 8: Find when a projectile lands

When a ball lands, its height equals zero. Factor the equation set to zero and apply the zero-product property, then reject any solution that gives a negative time, since time cannot be negative.

-16t^2 + 48t = 0 \Rightarrow -16t(t - 3) = 0 \Rightarrow t = 0 \text{ or } t = 3

$t = 0$ is the launch time; $t = 3$ is when it lands.

Final answer: The ball lands at $t = 3$ seconds.

Sources & how we know this

Georgia's K-12 Mathematics Standards (Algebra: Concepts & Connections) — Georgia Department of Education (2023)
Georgia Milestones Assessment System — Georgia Department of Education (2024)