Graphs of standard functions

(\sqrt{x^2}=|x|) for all real (x). In exams, remember modulus when square root and square appear together.

(x^2+1) is an even function and its minimum is (1), so it does not cut the (x)-axis. In exams, check both symmetry and minimum value.

The denominator (|x|) becomes zero at (x=0), so (0) is excluded. In exams, remove values that make the denominator zero.

The direct answer is option B, (-5,-2). The basic parabola y=x^2 has vertex (0,0). The vertex form y=(x-h)^2+k has vertex (h,k). Here x+5 can be written as x-(-5), so h=-5, and k=-2. Therefore the new vertex is (-5,-2). Equivalently, replacing x by x+5 shifts the graph 5 units left, while subtracting 2 shifts it 2 units down. Option B is correct. Option A, (5,-2), has the horizontal sign reversed; x+5 means left, not right. Option C, (-5,2), has the vertical sign reversed; -2 means downward. Option D, (2,-5), swaps the coordinate roles and does not match vertex form. A safe rule is x-h gives horizontal movement h, and the outside constant gives vertical movement directly. The supplied explanation and answer are correct.

(\lfloor2.3\rfloor=2) and (\lfloor-2.3\rfloor=-3), so the value is (5). In exams, avoid mistakes with greatest integers of negative numbers.

For (x>2), both (x+2) and (x-2) are positive, so both signum values are (1). In exams, check the sign of each signum term separately.

The denominator (x^2-9=(x-3)(x+3)) is zero at (x=\pm3). In exams, find vertical asymptotes from the zeroes of the denominator.

From (x+1\ge0), the domain is ([-1,\infty)), and the negative sign gives (y\le0). In exams, check both the square root and the outside negative sign.

(|x^2-4|=0) when (x^2-4=0), so (x=\pm2). In exams, modulus is zero only when the inside expression is zero.

The denominator (x+2=0) gives (x=-2), and the ratio of leading coefficients is (1). In exams, find the two asymptotes using different rules.

The modulus graph decreases up to the vertex (x=3) and then increases. In exams, remember the left arm of a (V)-graph is decreasing.

Let (t=|x|\ge0), then (f=t^2-2t=(t-1)^2-1). In exams, use (|x|) as a new variable to find the minimum.

In (\sqrt{x}=x-2), (x\ge2), and for (x=4) both sides are (2). In exams, check (x-2\ge0) before squaring.

The denominator (x^2+1) has minimum (1) at (x=0), so the fraction has maximum (1). In exams, a smaller positive denominator gives a larger fraction.

From (0=-|x-2|+4), (|x-2|=4), so (x=-2) or (x=6). In exams, add the distance in both directions after modulus.

Reflection in the (x)-axis gives (y=-x^2), and shifting (3) up gives (y=-x^2+3). In exams, track reflection and shift carefully.

(\frac{1}{x^2}>0) for (x\ne0), so the graph stays above and is undefined at (x=0). In exams, remember positive branches on both sides when (x^2) is in the denominator.

Between (-1) and (3), the total distance from both points remains constant (4). In exams, interpret the sum of two moduli as distance.

On ([2,3)), the value is (2), but (\lfloor3\rfloor=3). In exams, watch open and closed endpoints in a step graph.

Since (\sqrt{|{-x}|}=\sqrt{|x|}), the function is even. In exams, identify (y)-axis symmetry from (f(-x)=f(x)).

From (|x-1|+2=4), (|x-1|=2), so (x=-1) or (x=3). In exams, set the (y)-values equal for intersection with a horizontal line.

Between (0) and (2), the total distance has minimum (2), and it increases outside. In exams, use the distance idea for sums of moduli.

The minimum of (|x-4|) is (0) at (x=4), so (y=-2). In exams, set the inside of the modulus to (0) to find the vertex.

The parabola opens downward and the vertex (y)-value is (-1). In exams, the vertex of a downward parabola gives the maximum.

For the square root, (2x-5\ge0), so (x\ge\frac{5}{2}). In exams, keep the expression inside the square root non-negative.