Graphs of standard functions

Correct answer: A. domain ([3,\infty)) and range ((-\infty,2])

From (x-3\ge0), the domain is ([3,\infty)), and the negative sign gives (y\le2). In exams, check both the square root and the outside negative sign.

Correct answer: A. ((-3,0)) and ((3,0))

(|x^2-9|=0) when (x^2-9=0), so (x=\pm3). In exams, modulus is zero only when the inside expression is zero.

Correct answer: A. \(y=\frac{1}{x}\)

For \(y=1/x\), x and y have the same sign, so its branches lie in quadrants I and III. Neither x nor y can be zero, making both axes asymptotes. In contrast, \(1/x^2\) remains above the x-axis. Exam tip: use signs and powers to locate branches.

Correct answer: A. (x=5) and (y=3)

The denominator (x-5=0) gives (x=5), and the ratio of leading coefficients is (3). In exams, find the two asymptotes using different rules.

Correct answer: A. ([-4,\infty))

The modulus graph has vertex at (x=-4), and after that it increases. In exams, remember the right arm of a (V)-graph is increasing.

Correct answer: A. ( -4 )

Let (t=|x|\ge0), then (f=t^2-4t=(t-2)^2-4). In exams, use (|x|) as a new variable to find the minimum.

Correct answer: C. (x=3)

In (\sqrt{x+1}=x-1), (x\ge1), and at (x=3) both sides are (2). In exams, the right side must be non-negative before squaring.

Correct answer: A. (\frac{1}{2})

The denominator (x^2+4) has minimum (4) at (x=0), so the maximum is (\frac{2}{4}=\frac{1}{2}). In exams, a smaller denominator makes the fraction larger.

Correct answer: A. ((-8,0)) and ((2,0))

From (0=-|x+3|+5), (|x+3|=5), so (x=-8) or (x=2). In exams, add the distance in both directions after modulus.

Correct answer: B. (y=-|x|+2)

Reflection in the (x)-axis gives (y=-|x|), and shifting (2) up gives (y=-|x|+2). In exams, write reflection and shift separately.

Correct answer: A. (y=-\frac{1}{x})

In (y=-\frac{1}{x}), (x) and (y) have opposite signs. In exams, connect the negative reciprocal graph with the second and fourth quadrants.

Correct answer: A. ([2,8])

Between (2) and (8), the total distance from both points remains constant (6). In exams, interpret the sum of two moduli as distance.

Correct answer: A. left value near (-2) and value (-1) at the point

On ([-2,-1)), the value is (-2), but (\lfloor-1\rfloor=-1). In exams, watch open and closed points on a step graph.

Correct answer: A. (x=2)

In (|x-2|), points equally distant from (x=2) give the same (y). In exams, the symmetry line of a (|x-a|) graph is (x=a).

Correct answer: A. (x=-7) and (x=3)

From (|x+2|+1=6), (|x+2|=5), so (x=-7) or (x=3). In exams, set the (y)-values equal for intersection with a horizontal line.

Correct answer: A. ([4,\infty))

Between (-3) and (1), the total distance has minimum (4), and it increases outside. In exams, use the distance idea for sums of moduli.

Correct answer: A. (x=2) and (y=3)

The denominator ((x-2)^2) becomes zero at (x=2), and the outside (3) gives horizontal asymptote (y=3). In exams, exclude the value that makes a squared denominator zero.

Correct answer: A. ( (2,3) )

The direct answer is A, \((2,3)\). The basic graph \(y=|x|\) has its vertex at \((0,0)\). In \(y=|x-2|+3\), replacing x by x−2 shifts the graph 2 units right, and adding 3 outside shifts it 3 units up. Therefore the vertex becomes \((2,3)\). Equivalently, the absolute value is smallest when x−2=0, so x=2; then y=|0|+3=3. Option A is correct. Option B, \((-2,3)\), would result from \(|x+2|+3\), a left shift. Option C, \((3,2)\), reverses the coordinate roles. Option D, \((0,3)\), ignores the horizontal shift. Memory cue: in \(|x-h|+k\), the vertex is \((h,k)\), and the graph is a V opening upward.

Correct answer: A. (x=3)

The denominator becomes zero when (x-3=0), so (x=3). For reciprocal graphs, set the denominator equal to zero.

Correct answer: A. ( -4 )

The minimum of ((x+1)^2) is (0), hence (y_{\min}=-4). A completed square often gives the minimum directly.

Correct answer: A. (y\le2)

The negative sign makes the graph open downward, and the maximum value is (2). For range, check the vertex and opening direction.

Correct answer: A. (y=(x-2)^3+5)

A right shift replaces (x) by (x-2), and an upward shift adds (5). In transformations, the inside sign works oppositely.

Correct answer: A. (2)

For the greatest integer function, the value is (2) on (2\le x<3). Its step graph has a closed left end and open right end.

Correct answer: A. 0 ≤ y < 1

The governing concept is the fractional-part function. For every real number x, its fractional part is defined by {x}=x−⌊x⌋, where ⌊x⌋ is the greatest integer not exceeding x. Consequently, 0≤{x}<1. The value 0 occurs whenever x is an integer, while 1 is never reached because the integer portion would then increase by one. Therefore the range is [0,1), represented by 0≤y<1, so option A is correct. Option B excludes the attainable value 0 and includes the impossible value 1. Option C confuses fractional values with integers, and option D is too broad because values greater than or equal to 1 cannot occur.

Correct answer: A. (2)

Solving (x^2=2x+3) gives (x^2-2x-3=0), which has two real roots. Intersections equal the number of real solutions.