﻿ Linear and Quadratic Functions Worksheet | Problems & Solutions

# Linear and Quadratic Functions Worksheet

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1.
A small stone is projected directly upward from the ground with an initial velocity of 32 ft/s. Write an equation to find the time $t$ for the stone to reach a height of 64 ft.
 a. 16$t$2 + 32$t$ + 64 = 0 b. 32$t$2 + 16$t$ - 64 = 0 c. 16$t$2 + 64$t$+ 32 = 0 d. 16$t$2 - 32$t$ + 64 = 0

#### Solution:

The velocity of projection is u = 32 ft/s

The height reached by the vertically projected stone in t sec is h = ut - 16t2
[Use the laws of motion for a vertically projected body.]

64 = 32t - 16t2
[Replace h = 64 ft and u = 32 ft/s to find the time for the stone to reach the height 64 ft.]

16t2 - 32t + 64 = 0

2.
In a zoo, there are some apes. 4 apes are on the trees and the number of remaining apes is equal to the square of one-fifth of their total number. Choose the total number of apes from the following.
 a. 20 b. 16 c. 4 d. 10

#### Solution:

Total number of apes = 4 + (1 / 5 of total number of apes)2
[Aanalyze the situation.]

n = 4 + n²25
[Let the total number of apes = n.]

n2 - 25n + 100 = 0
[Multiply each term with 25 and rearrange.]

(n - 20)(n - 5) = 0
[Factor.]

n - 20 = 0 or n - 5 = 0

n =20 or n = 5
[Solve for n.]

The number of apes in the forest is either 20 or 5.

3.
There are three consecutive positive integers. If the product of the first two integers is 2 more than 9 times the third integer, then find the integers.
 a. 2, 3, 4 b. 8, 9, 10 c. 10, 11, 12 d. 6, 7, 8

#### Solution:

The consecutive integers are x, x + 1, x + 2
[Let the first positive integer be x.]

x(x + 1) = 2 + 9(x + 2)
[Analyze the situation.]

x2 + x = 2 + 9x + 18
[Use the Distributive property.]

x2 - 8x - 20 = 0
[Group the terms and simplify.]

(x - 10)(x + 2) = 0
[Factor.]

x = 10 or x = - 2
[Solve for x.]

x = 10
[x is positive.]

The consecutive positive integers are 10, 11, 12.

4.
Find the sides of a right triangle which are consecutive integers.
 a. 5, 6, 7 b. cannot be determined c. 3, 4, 5 d. 2, 3, 4

#### Solution:

The sides of the right triangle are x, x + 1, and x + 2
[Let the short leg of the right triangle = x, and analyze the situation.]

x2 + (x + 1)2 = (x + 2)2
[Use the Pythagorean theorem.]

x2 + x2 + 2x + 1 = x2 + 4x + 4
[Use (a + b)2 = a2 + 2ab + b2, and expand.]

x2 - 2x - 3 = 0
[Group the terms and simplify.]

(x - 3)(x + 1) = 0
[Factor.]

x = 3 or x = - 1
[Solve for x.]

x = 3
[The length of the side of a triangle cannot be negative.]

The sides of the right triangle are 3 units, 4 units, and 5 units.

5.
In a zoo, there are some apes. 6 apes are on the trees and the number of remaining apes is equal to the square of one-seventh of their total number. Choose the total number of apes from the following.
 a. 42 b. 41 c. 49 d. 43

#### Solution:

Total number of apes = 6 + (1 / 7 of total number of apes)2
[Aanalyze the situation.]

n = 6 + n²49
[Let the total number of apes = n.]

n2 - 49n + 294 = 0
[Multiply each term with 49 and rearrange.]

(n - 42)(n - 7) = 0
[Factor.]

n - 42 = 0 or n - 7 = 0

n =42 or n = 7
[Solve for n.]

The number of apes in the forest is either 42 or 7.

6.
Find the rate of change of the linear function $y$ = 7$x$ + 10.
 a. 10 b. - 10 c. 7 d. - 7

#### Solution:

The rate of change of the linear function y = 7x + 10 is 7.
[The rate of change of the linear function y = mx + b is m.]

7.
Find the leading coefficient of the given function ($p$) = 29.
 a. 29 b. 1 c. 2

#### Solution:

The leading coefficient of a polynomial function is the coefficient of highest degree variable in it.

Here, the polynomial function is (p) = 29.

(p) = 29 · p0
[Rewrite (p) using p0 = 1.]

So, the leading coefficient of (p) is 29.

8.
What is the lower bound of the correlation coefficient $r$?
 a. 1 b. - 2 c. - 1

#### Solution:

The correlation coefficient r [- 1, 1] and hence its lower bound is -1.

9.
What is the upper bound the correlation coefficient $r$?
 a. 1 b. 2 c. -1

#### Solution:

The correlation coefficient r [- 1, 1] and hence its upper bound is 1.

10.
Write the degree of the polynomial function ($v$) = 8$v$11 + 46.
 a. 11 b. 8 c. 13

#### Solution:

The degree of the polynomial function f (x) = an xn + an-1 xn-1 + ... + a2 x2 + a1 x + a0 is n.

So, the degree of the polynomial function (v) = 8v11 + 46 is 11.