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

Use the axis of symmetry x equals negative b over 2a to get the x-coordinate, then substitute it back into the function for the y-coordinate. For f of x equals x squared minus 6x plus 5, x equals 6 over 2 equals 3, and f of 3 equals negative 4, so the vertex is the point (3, negative 4). This formula is not on the reference sheet, so memorize it.

What does each form of a quadratic reveal?

Standard form, ax squared plus bx plus c, reveals the y-intercept c. Vertex form, a times the quantity x minus h squared plus k, reveals the vertex (h, k). Factored form, a times (x minus p)(x minus q), reveals the zeros p and q. All three share the same leading coefficient a, which sets the direction of opening and width.

What are the four ways to solve a quadratic on the EOC?

Factoring with the zero-product property, taking square roots, completing the square, and the quadratic formula. Factoring is fastest when integer factors exist. Square roots suit a squared binomial. Completing the square works on any quadratic and builds vertex form. The quadratic formula, on the reference sheet, solves every quadratic and is the safe fallback.

What does the discriminant tell you?

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

How do you complete the square?

Move the constant to the other side, then add the square of half the linear coefficient, that is (b over 2) squared, to both sides. The left side becomes a perfect-square trinomial that factors as a binomial squared. Take the square root of both sides, keeping the plus-or-minus, and solve. If the leading coefficient is not 1, divide every term by it first.

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

The vertex answers the extreme value: maximum height, maximum revenue, or minimum cost, reporting the y-coordinate as the value. The zeros answer when something is at ground level or breaks even, such as a projectile's landing time. Words like how high or maximum point to the vertex, while when does it land points to a zero. Reject any solution that cannot happen, like a negative time or length.

FloridaMaths

B.E.S.T. Algebra 1 EOC: a complete guide to quadratic functions and equations

A deep-dive B.E.S.T. Algebra 1 EOC guide to quadratic functions and equations (MA.912.AR.3): graphing parabolas and key features, the three forms, solving by factoring, square roots, completing the square, and the quadratic formula, the discriminant, and applications. Where the higher achievement levels are won.

Generated by Claude Opus 4.817 min readMA.912.AR.3Updated 2026-06-02

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

Jump to a section

What this category demands
Graphing and key features
The three forms
Solving by factoring, square roots, and completing the square
The quadratic formula and the discriminant
Applications
How this category is examined
Check your knowledge

What this category demands

This guide covers quadratic functions and equations (MA.912.AR.3), which straddle the Functions and Modeling and Algebra and Modeling categories and together account for a large share of the B.E.S.T. Algebra 1 EOC. Quadratics are usually where the Level 4 and Level 5 standards are decided, so mastering them lifts a passing score into mastery. Each dot-point page has its own practice: graphing quadratic functions, forms of a quadratic, solving by factoring, solving by square roots and completing the square, the quadratic formula and the discriminant, and quadratic applications.

Graphing and key features

A quadratic graphs as a parabola. The sign of $a$ sets the direction: $a > 0$ opens up (minimum), $a < 0$ opens down (maximum). The axis of symmetry and vertex $x$ -coordinate is $x = \frac{-b}{2a}$ (memorize, not on the reference sheet); substitute for the vertex's $y$ . The $y$ -intercept is $f(0) = c$ , and the $x$ -intercepts are the solutions of $f(x) = 0$ . Because the on-screen calculator is scientific, compute all of these from the equation.

The three forms

Standard $ax^2 + bx + c$ shows the $y$ -intercept; vertex $a(x - h)^2 + k$ shows the vertex $(h, k)$ ; factored $a(x - p)(x - q)$ shows the zeros. Convert to standard by expanding, to factored by factoring, and to vertex by completing the square. Watch signs: $(x + 3)^2$ has vertex $x = -3$ .

Solving by factoring, square roots, and completing the square

To factor, set to zero and use the zero-product property; solutions are the zeros. 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 both sides to make a perfect square, solving any quadratic and producing vertex form.

The quadratic formula and the discriminant

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

Applications

Quadratics model projectile motion ( $h(t) = -16t^2 + v_0 t + h_0$ , ground at $h = 0$ , peak at the vertex) and area (length times width). Solve, then interpret: the vertex is a maximum or minimum, the zeros are start and end points. Reject negative times, lengths, or counts and state units.

A full projectile problem

A rocket's height is $h(t) = -16t^2 + 80t + 6$ feet. Find the maximum height and the landing time (to the nearest tenth).

step 1 Maximum at the vertex

$t = \frac{-80}{2(-16)} = 2.5$ s; $h(2.5) = -16(6.25) + 80(2.5) + 6 = -100 + 200 + 6 = 106$ feet.

step 2 Landing where height is zero

$-16t^2 + 80t + 6 = 0$ ; quadratic formula with $a = -16$ , $b = 80$ , $c = 6$ .

step 3 Apply and choose the realistic root

$t = \frac{-80 \pm \sqrt{6400 + 384}}{-32} = \frac{-80 \pm \sqrt{6784}}{-32}$ , $\sqrt{6784} \approx 82.4$ ; the positive root is $t \approx 5.1$ s. Maximum 106 feet; lands at about 5.1 seconds.

How this category is examined

Equation editor and number entry. Solve by any method (simplest radical form), find a vertex, or compute a maximum or landing time.
GRID and hot-spot. Plot the vertex or intercepts.
Multiple choice and multiselect. Count solutions from the discriminant, identify a form's feature, or select true statements.
Context items. Model and interpret projectile and area problems.

Check your knowledge

Work these as you would for credit on the computer-based test.

Find the vertex of $f(x) = x^2 - 8x + 11$ . (2 points)
State the zeros of $f(x) = (x - 3)(x + 7)$ . (1 point)
Solve $x^2 - 5x - 14 = 0$ by factoring. (2 points)
Solve $(x + 1)^2 = 16$ . (1 point)
Solve $x^2 + 6x + 2 = 0$ by completing the square (simplest radical form). (2 points)
Solve $3x^2 + 2x - 1 = 0$ with the quadratic formula. (2 points)
How many real solutions does $x^2 + x + 5 = 0$ have? (1 point)
A ball's height is $h(t) = -16t^2 + 48t$ . When does it land? (2 points)
A rectangle is 4 longer than wide with area 45. Find the width. (2 points)

Solutions

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

The $x$ -coordinate of the vertex is $x = \frac{-b}{2a}$ , which also locates the axis of symmetry. Substitute that $x$ -value back into the function to find the $y$ -coordinate. Memorize this formula because it is not on the reference sheet.

$a = 1$ , $b = -8$ : $x = \frac{-(-8)}{2(1)} = \frac{8}{2} = 4$ .

f(4) = 4^2 - 8(4) + 11 = 16 - 32 + 11 = -5

Final answer: Vertex $(4, -5)$ .

Step 2: Read the zeros from factored form

Factored form $a(x - p)(x - q)$ reveals the zeros directly: each factor set to zero gives a solution. Watch the signs carefully because the zero is opposite to the sign inside the parentheses.

$(x - 3)(x + 7) = 0 \implies x - 3 = 0$ or $x + 7 = 0$ .

Final answer: $x = 3$ and $x = -7$ .

Step 3: Solve a quadratic by factoring and the zero-product property

Move all terms to one side, factor the trinomial, then apply the zero-product property: if a product of factors equals zero, at least one factor must be zero. Find two numbers that multiply to $c$ and add to $b$ .

Find factors of $-14$ that add to $-5$ : $-7$ and $2$ .

(x - 7)(x + 2) = 0 \implies x = 7 \text{ or } x = -2

Final answer: $x = 7$ or $x = -2$ .

Step 4: Solve a squared-binomial equation using the square-root property

When a perfect square equals a constant, take the square root of both sides and include both the positive and negative cases. The $\pm$ symbol is essential because both roots are valid.

(x + 1)^2 = 16 \implies x + 1 = \pm 4 \implies x = 3 \text{ or } x = -5

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

Step 5: Solve a quadratic by completing the square

Move the constant to the right, then add $\left(\frac{b}{2}\right)^2$ to both sides to create a perfect-square trinomial on the left. Take the square root of both sides, include $\pm$ , and solve for $x$ . This method works for any quadratic.

Move constant: $x^2 + 6x = -2$ . Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides: $(x + 3)^2 = 7$ .

x + 3 = \pm\sqrt{7} \implies x = -3 \pm \sqrt{7}

Final answer: $x = -3 \pm \sqrt{7}$ .

Step 6: Solve a quadratic using the quadratic formula

The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is on the reference sheet and solves every quadratic. Identify $a$ , $b$ , and $c$ carefully, paying attention to signs, then simplify the discriminant and the resulting fractions.

$a = 3$ , $b = 2$ , $c = -1$ : discriminant $= 4 + 12 = 16$ .

x = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \implies x = \frac{2}{6} = \frac{1}{3} \text{ or } x = \frac{-6}{6} = -1

Final answer: $x = \frac{1}{3}$ or $x = -1$ .

Step 7: Count real solutions using the discriminant

The discriminant $b^2 - 4ac$ tells you how many real solutions exist without fully solving. A negative discriminant means the square root is imaginary, so there are no real solutions.

$b^2 - 4ac = 1^2 - 4(1)(5) = 1 - 20 = -19 < 0$ .

Final answer: No real solutions.

Step 8: Find when a projectile lands by solving for zero height

Set the height function equal to zero and solve. Factor out common factors first to simplify the equation. Reject $t = 0$ as the launch time and report only the positive root as the landing time.

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

$t = 0$ is the launch; $t = 3$ is the landing.

Final answer: Lands at $t = 3$ seconds.

Step 9: Solve an area word problem using a quadratic equation

Assign a variable to the unknown dimension, express the other dimension in terms of it, and set up the area equation. Set it equal to zero, factor, and reject any negative solution because a length cannot be negative.

Let $w$ = width; then length $= w + 4$ .

w(w + 4) = 45 \implies w^2 + 4w - 45 = 0 \implies (w + 9)(w - 5) = 0

$w = -9$ is rejected (negative length); $w = 5$ .

Final answer: Width $= 5$ .

Sources & how we know this

B.E.S.T. Algebra 1 EOC Reference Sheet — Florida Department of Education (2024)
B.E.S.T. Mathematics Standards — Florida Department of Education (2020)