Algebra 2 Forms of Quadratic Functions Worksheet Solutions (PDF)

Algebra 2 math worksheet solutions: Forms of Quadratic Functions – Maths PDF

Cover of the book Algebra 2 Math Worksheets with Answers (PDF)

Download as a PDFThe Algebra 2 worksheets book with answersPrintable PDFSee the book ›

Sommaire

This quadratic forms answers for Algebra 2 covers forms of quadratic functions (Common Core A-SSE.B.3a, A-SSE.B.3b, F-IF.C.8a), with a free PDF to print.

Every worksheet problem on the forms of quadratic functions is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where a sign error when finding the vertex or the number to complete the square can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: A-SSE.B.3a, A-SSE.B.3b, F-IF.C.8a

Quadratic forms answers: exercises 1 to 10

Exercise 1 – Read the y-intercept

The y-intercept is $f(0)$, which is the constant term $c$.

$(0, 7)$

Exercise 2 – Zeros and vertex from factored form

Zeros: $x = 3$ and $x = -1$.

The axis is halfway between them: $x = \frac{3 + (-1)}{2} = 1$.

$f(1) = 2(-2)(2) = -8$.

Zeros 3 and -1; axis x = 1; vertex (1, -8)

Exercise 3 – Read vertex form

Compare with $a(x – h)^{2} + k$: $h = -2$, $k = 5$, $a = -1$.

Vertex (-2, 5); maximum value 5 because a < 0

Exercise 4 – Vertex from standard form

$x = -\frac{-8}{2(1)} = 4$. $f(4) = 16 – 32 + 11 = -5$.

$(4, -5)$

Exercise 5 – Factor to find zeros

Find two numbers with product $-14$ and sum $-5$: $-7$ and $2$.

$(x – 7)(x + 2) = 0$.

$x = 7, -2$

Exercise 6 – Factor with a leading coefficient

$3x^{2} + 10x – 8 = (3x – 2)(x + 4)$. Check: $3x^{2} + 12x – 2x – 8$.

$x = \frac{2}{3}, -4$

Exercise 7 – Complete the square, area model

Corner 25.

Half of 10 is 5, and $5^{2} = 25$.

$x^{2} + 10x + 25 = (x + 5)^{2}$

Exercise 8 – Complete the square

Take half of the $x$-coefficient $-10$: $-5$. Square it: $25$.
Add and subtract $25$: $x^{2} – 10x + 3$ becomes $(x^{2} – 10x + 25) + 3 – 25$.
Write the perfect square and combine constants: $y = (x – 5)^{2} – 22$.

$(x – 5)^{2} – 22$

Exercise 9 – Complete the square, a = 2

Factor $2$ from the $x^{2}$ and $x$ terms: $2(x^{2} + 6x) + 7$.
Take half of the $x$-coefficient $6$: $3$. Square it: $9$.
Add and subtract $9$ inside the parentheses: the subtracted amount becomes $2 \cdot 9 = 18$ outside.
Write the perfect square and combine constants: $y = 2(x + 3)^{2} – 11$.

$2(x + 3)^{2} – 11$; vertex $(-3, -11)$

Exercise 10 – Odd coefficient

Take half of the $x$-coefficient $3$: $\frac{3}{2}$. Square it: $\frac{9}{4}$.
Add and subtract $\frac{9}{4}$: $x^{2} + 3x + 1$ becomes $(x^{2} + 3x + \frac{9}{4}) + 1 – \frac{9}{4}$.
Write the perfect square and combine constants: $y = (x + \frac{3}{2})^{2} – \frac{5}{4}$.

$(x + \frac{3}{2})^{2} – \frac{5}{4}$

Exercises 11 and up

Exercise 11 – Vertex and axis

Take half of the $x$-coefficient $4$: $2$. Square it: $4$.
Add and subtract $4$: $x^{2} + 4x + 9$ becomes $(x^{2} + 4x + 4) + 9 – 4$.
Write the perfect square and combine constants: $y = (x + 2)^{2} + 5$.

Vertex $(-2, 5)$; axis $x = -2$; minimum $5$

Exercise 12 – Equation from the graph, vertex

y = (x – 3)² – 2.

$f(x) = a(x – 3)^{2} – 2$. Substitute $(5, 2)$: $2 = a(4) – 2$, so $a = 1$.

$f(x) = (x – 3)^{2} – 2$

Exercise 13 – Equation from zeros

$f(x) = a(x + 1)(x – 5)$. At $x = 0$: $a(1)(-5) = -5$, so $a = 1$.

$f(x) = (x + 1)(x – 5)$

Exercise 14 – Maximum value

$x = -\frac{8}{2(-2)} = 2$. $f(2) = -8 + 16 + 1 = 9$.

Since $a < 0$, this is a maximum. Vertex form: $-2(x - 2)^{2} + 9$.

Maximum 9 at x = 2

Exercise 15 – Thrown ball

Vertex (2, 69).

$-16(t^{2} – 4t) + 5 = -16(t^{2} – 4t + 4) + 64 + 5 = -16(t – 2)^{2} + 69$.

Maximum height 69 feet at t = 2 seconds

Exercise 16 – Revenue

Vertex form: $-5(p^{2} – 20p) = -5(p^{2} – 20p + 100) + 500 = -5(p – 10)^{2} + 500$.

Factored form: $-5p(p – 20)$, zeros 0 and 20.

Price $10 gives maximum revenue $500

Exercise 17 – Fence

Zeros: $w = 0$ and $w = 40$. The vertex is halfway: $w = 20$.

$A(20) = 20 \cdot 20 = 400$.

20 ft by 20 ft; area 400 square feet

Exercise 18 – Show two forms are equal

Left: $x^{2} – 6x + 9 – 4 = x^{2} – 6x + 5$.

Right: $x^{2} – 5x – x + 5 = x^{2} – 6x + 5$.

Both equal $x^{2} – 6x + 5$

Exercise 19 – Which form?

Vertex: $(2, -9)$. Zeros: $(x – 2)^{2} = 9$, so $x = 2 \pm 3 = 5, -1$.

y-intercept: $f(0) = 4 – 9 = -5$.

Vertex (2, -9); zeros 5 and -1; y-intercept (0, -5)

Exercise 20 – Find the error

$(x + 4)^{2} = x^{2} + 8x + 16$, which adds 16. He must subtract 16: $x^{2} + 8x + 3 = (x + 4)^{2} – 16 + 3$.

$(x + 4)^{2} – 13$

Exercise 21 – No x-intercepts

A square is never negative, so $(x – 2)^{2} \ge 0$ and $f(x) \ge 3$. The lowest point is $(2, 3)$, which is above the x-axis, and the parabola opens up.

The minimum value 3 is above 0

Key facts behind this quadratic forms answers

Key facts

Standard form ax² + bx + c: y-intercept c.
Factored form a(x – r)(x – s): zeros r and s.
Vertex form a(x – h)² + k: vertex (h, k) and extreme value k.
Complete the square: add and subtract (b/2a)².
a > 0 opens up (minimum); a < 0 opens down (maximum).

How to use this quadratic forms answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the quadratic forms answers as a PDF with the button below and keep it with your notes.

All Algebra 2 chapters: Algebra 2 math.

Voter.. post

Free printable PDF