Graphs of standard functions

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.

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

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.

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.

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.

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.

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.

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.

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

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

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

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

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

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).

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.

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

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.

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.

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

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

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

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

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.

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.

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