Graphs of standard functions

The denominator (x+2) is zero at (x=-2), and the outside (3) gives horizontal asymptote (y=3). In exams, check both denominator and vertical shift.

For (x\ge5), (|x-5|=x-5) and (|x+1|=x+1), so (y=-6). In exams, open modulus according to the interval.

When (x\in[-3,-2)), (x+1\in[-2,-1)), so (\lfloor x+1\rfloor=-2). In exams, first transform the inside interval.

The direct answer is option A, -1. The signum function returns 1 when its input is positive, 0 when its input is zero, and -1 when its input is negative. If x>2, subtracting x from 2 gives 2-x<0. Therefore sgn(2-x)=-1, so the graph value is -1. Option A is correct. Option B, 0, would occur only when 2-x=0, which means x=2, not x>2. Option C, 1, would occur when 2-x>0, meaning x<2. Option D, 2, is the constant appearing inside the expression, not a possible signum output in this situation; signum outputs only -1, 0, or 1. The boundary x=2 is important: the value there is 0, while all values strictly to the right give -1. The supplied answer is consistent. Remember to determine the sign of the entire inside expression before applying signum.

Every point on the \(x\)-axis has \(y=0\). Hence, \(3|x+2|-9=0\), which gives \(|x+2|=3\). Taking both cases, \(x+2=3\) gives \(x=1\), while \(x+2=-3\) gives \(x=-5\). Therefore, the intercepts occur at \(x=-5\) and \(x=1\). Option B does not result from solving the modulus equation correctly. Exam tip: for \(|u|=a\) with \(a>0\), always consider both equations \(u=a\) and \(u=-a\).

Since (\sqrt{x+4}\ge0), (5-\sqrt{x+4}\le5). In exams, a negative square-root term makes the graph go downward.

(x+3) shifts (3) units left, outside (2) vertically stretches, and (-1) shifts down. In exams, read inside and outside transformations separately.

Equating gives (x^2-1=2x+2), that is (x^2-2x-3=0). In exams, set the two (y)-values equal for intersections.

The denominator (x-4) becomes zero at (x=4). In exams, find the vertical asymptote of a reciprocal graph from the denominator.

From (3x-12=0), (x=4) and the minimum is (0). In exams, find the vertex by setting the inside of the modulus to (0).

In this interval, \(2x\) starts in ([3,4)) and also reaches values giving \(4\) after \(2x\ge4\). In exams, split the inside interval into subintervals.

Squaring gives (x^2+y^2=16) with (y\ge0). In exams, identify (\sqrt{r^2-x^2}) as an upper semicircle.

For \(x>0\), \(|x|=x\), so \(\frac{|x|}{x}=1\). For \(x<0\), \(|x|=-x\), so \(\frac{|x|}{x}=-1\). Therefore, its graph is the same as \(y=\operatorname{sgn}(x)\) on the domain \(x\ne0\); the function is not defined at \(x=0\). Option B reverses both values, while options C and D vary continuously with \(x\). Exam tip: split expressions containing \(|x|\) into the cases \(x>0\) and \(x<0\).

The minimum of (\sqrt{x-2}) is (0) at (x=2), so (y=4). In exams, a negative square-root graph has maximum at the starting point.

The axis of symmetry is (x=-\frac{b}{2a}=-\frac{-12}{4}=3). In exams, use the axis formula for a quadratic graph.

The degrees of numerator and denominator are equal, so the ratio of leading coefficients is (\frac{5}{2}). In exams, use the ratio of leading coefficients for equal degrees.

(\lfloor3.8\rfloor-2=1), so ((3.8,0)) is not on the graph. In exams, first find the greatest integer of (x).

The distance between (7) and (2) is (5), so the minimum sum is (5). In exams, remember the minimum of (|x-a|+|x-b|) is (|a-b|).

The minimum of (|x-2|) is (0) at (x=2), so (y=9) is maximum. In exams, a modulus graph with a negative multiplier has maximum at the vertex.

For \(y=x^2+2\), \(f(-x)=(-x)^2+2=x^2+2=f(x)\), so it is an even function and its graph is symmetric about the y-axis. Since \(x^2\geq 0\), we have \(y=x^2+2\geq 2\); hence the minimum value is 2, attained at \(x=0\). Option C has minimum value 0 and is symmetric about \(x=2\), not the y-axis. Exam tip: test y-axis symmetry using \(f(-x)=f(x)\).

The denominator (|x-1|) becomes zero at (x=1), so (1) is excluded. In exams, remove values that make the denominator zero.

(\lfloor1.7\rfloor=1) and (\lfloor-1.7\rfloor=-2), so the value is (-1). In exams, avoid mistakes with greatest integers of negative numbers.

For (1<x<4), (\operatorname{sgn}(x-1)=1) and (\operatorname{sgn}(x-4)=-1). In exams, check the sign of each signum term separately.

The denominator (x^2-4x+3=(x-1)(x-3)) is zero at (x=1) and (x=3). In exams, find vertical asymptotes from the zeroes of the denominator.

The direct answer is option C: the constant value is 3. When x\ge4, both x-1 and x-4 are non-negative, so |x-1|=x-1 and |x-4|=x-4. Substitute these forms: y=(x-1)-(x-4)=x-1-x+4=3. Therefore the expression is constant throughout the interval x\ge4. Option C is correct. Option A has the wrong sign. Option B would mean the two absolute values cancel completely, but their difference is 3. Option D, 2x-5, results from an incorrect sign expansion and is not constant. Memory cue: first identify the interval, then remove each modulus with the correct sign.