Graphs of standard functions

For \(x>5\), both \(x+1\) and \(x-5\) are positive. Hence, \(|x+1|=x+1\) and \(|x-5|=x-5\). Therefore, \(y=(x+1)-(x-5)=6\), so option A is correct. Exam tip: check the sign of each expression before removing modulus bars; \(2x-4\) would result from adding the terms instead of subtracting them.

The quadratic function is \(y=-x^2+4x+5\). Completing the square gives \(y=-(x-2)^2+9\). Since \((x-2)^2\geq 0\), we have \(-(x-2)^2\leq 0\), so the maximum value of \(y\) is \(9\), attained at \(x=2\). Option 5 is only the value at \(x=0\), not the maximum. Exam tip: for a downward-opening parabola, the y-coordinate of the vertex is the maximum value.

Every point on the \(x\)-axis has \(y=0\). Therefore, \(x^2+2x-8=0\), which factors as \((x+4)(x-2)=0\). Hence, \(x=-4\) or \(x=2\). Option B reverses the signs of both roots. Exam tip: To find the \(x\)-intercepts of a graph, set \(y=0\).

The endpoint of a square-root graph occurs when the expression under the radical becomes zero. Thus, \(x+9=0\), so \(x=-9\), and \(y=\sqrt{0}-4=-4\). Therefore, the endpoint is \((-9,-4)\). Option B reverses the horizontal shift. Exam tip: for \(y=\sqrt{x-h}+k\), the endpoint is \((h,k)\).

As \(|x|\) becomes very large, \(\frac{4}{x+2}\) approaches \(0\). Therefore, \(y\) approaches \(-5\), so the horizontal asymptote is \(y=-5\). The line \(x=-2\) is the vertical asymptote, not the horizontal one. Exam tip: For \(y=\frac{a}{x-h}+k\), the horizontal asymptote is directly \(y=k\).

The numerator and denominator are both of degree 1. Therefore, the horizontal asymptote is determined by the ratio of the leading coefficients: \(y=\frac{1}{1}=1\). Hence, \(y=1\) is correct. The line \(x=2\) is the vertical asymptote because the denominator becomes zero there. Exam tip: For a rational function with equal numerator and denominator degrees, divide the leading coefficient of the numerator by that of the denominator to find the horizontal asymptote.

The direct answer is option A: a hole. Factor the numerator: \(x^2-9=(x-3)(x+3)\). For \(x\ne3\), the expression simplifies to \(\frac{(x-3)(x+3)}{x-3}=x+3\). However, the original denominator is zero at \(x=3\), so that input is excluded even after cancellation. The simplified line would approach the missing point \((3,6)\), producing a removable discontinuity, or hole. Option A is correct. Option B, a vertical asymptote, would occur when the denominator’s zero factor remains uncancelled, as in \(1/(x-3)\). Option C, a vertex, belongs to a parabola or similar turning graph, not this cancelled line. Option D, a maximum point, is not present because the line has no maximum. Always check the original domain after cancelling factors; cancellation does not restore a forbidden input.

The domain is all real numbers because \(x^2+1\neq 0\). Also, \(f(-x)=\frac{(-x)^3}{(-x)^2+1}=-\frac{x^3}{x^2+1}=-f(x)\). Therefore, the function is odd, so its graph is symmetric about the origin. It is not symmetric about the y-axis, since that would require \(f(-x)=f(x)\). Exam tip: the condition \(f(-x)=-f(x)\) indicates symmetry about the origin.

The greatest integer function \(\lfloor x\rfloor\) gives the greatest integer less than or equal to \(x\). Since \(-4\) is itself an integer, \(\lfloor -4\rfloor=-4\), so \(f(-4)=-4\). The value \(-5\) is incorrect because it is less than \(-4\); in exams, remember that the greatest integer function of any integer is the integer itself.

The direct answer is option A:
\(g(-2.2)=-2\). The ceiling function gives the smallest integer that is greater than or equal to the given number. On the number line, \(-3<-2.2<-2\). Therefore, \(-2\) is the first integer at or above \(-2.2\), even though it is numerically less negative. Option A is correct because \(-2\ge -2.2\). Option B, \(-3\), is smaller than \(-2.2\), so it cannot be the ceiling. Option C, \(2\), changes the sign and is not the nearest required integer. Option D, \(3\), is also positive and too large. Do not confuse ceiling with rounding: \(-2.2\) rounds to \(-2\), but the ceiling rule is based on the least integer not below the number. Memory cue: ceiling means move upward on the number line, including for negative numbers.

The signum function \(\operatorname{sgn}(t)\) equals \(0\) only when \(t=0\). Hence, solve \(x^2-4=0\), or \((x-2)(x+2)=0\), giving \(x=2\) and \(x=-2\). Option C is incorrect because \(4^2-4\neq0\). Exam tip: To find where \(\operatorname{sgn}(g(x))=0\), set the inner expression \(g(x)\) equal to zero.

The function can be written as \(y=3\) for \(x\leq 3\) and \(y=x\) for \(x\geq 3\). These two lines meet at \(x=3\), where \(y=3\), so the corner is at \((3,3)\). The point \((0,3)\) lies on the horizontal part but is not a point where the slope changes. Exam tip: the graph of \(\max(x,a)\) has its corner at \((a,a)\).

For \(x\leq -3\), the minimum is \(x\), so \(y=x\); for \(x\geq -3\), the minimum is \(-3\), so \(y=-3\). These two rules meet at \(x=-3\), giving \(y=-3\). Hence the corner is \((-3,-3)\). Option D lies on the horizontal part of the graph but is not the point where the rule changes. Exam tip: find the intersection of \(y=x\) and \(y=-3\).

When \(x\ge 2\), we have \(x-2\ge 0\), so \(|x-2|=x-2\). Therefore, \(y=(x-2)+x=2x-2\), making \(y=2x-2\) the correct line. Option D represents only the modulus term \(|x-2|\) and omits the added \(+x\). Exam tip: when the expression inside a modulus is non-negative, its modulus equals the expression itself.

At an intersection point, both graphs have the same \(y\)-value, so set \(x^2=3x+4\). Rearranging and factoring gives \(x^2-3x-4=0\), or \((x-4)(x+1)=0\). Hence \(x=4\) or \(x=-1\), making option A correct. Option B results from taking the sign of the root \(-1\) incorrectly. Exam tip: to find intersections of two graphs, equate their \(y\)-expressions first.

At an intersection, the two graphs have the same \(y\)-value, so \(|x-2|=4\). Thus \(x-2=4\) or \(x-2=-4\), giving \(x=6\) or \(x=-2\). Therefore, option A is correct. Option B does not satisfy \(|x-2|=4\). Exam tip: when \(|u|=a\) with \(a>0\), solve both cases \(u=a\) and \(u=-a\).

At the intersection, the two graphs have the same \(y\)-value, so \(\sqrt{x+1}=3\). Squaring both sides gives \(x+1=9\), hence \(x=8\). The distractor \(x=9\) results from forgetting to subtract 1. Exam tip: Always substitute the obtained value back into the original equation; here, \(\sqrt{8+1}=3\).

In \(f(x+4)\), the input contains \(+4\), so the graph shifts 4 units horizontally to the left; the sign inside the function reverses the direction. The outside term \(-3\) subtracts 3 from every function value, shifting the graph 3 units downward. Therefore, option A is correct. Exam tip: \(f(x-a)\) shifts a graph a units right, while \(f(x)+b\) shifts it b units up.

In \(y=2f(x)-1\), the factor 2 outside \(f(x)\) doubles every \(y\)-value, producing a vertical stretch by a factor of 2. The term \(-1\) then shifts the entire graph 1 unit downward. It is not a horizontal stretch, because horizontal transformations involve changes inside the function’s argument, such as \(f(2x)\). Exam tip: factors outside the function affect vertical changes, while factors inside its argument affect horizontal changes.

With \(\frac{x}{2}\) inside, (x) must double to get the same (f)-value. Hence the graph stretches horizontally.

Replacing x by \(-x\) in \(f(-x)\) reflects the original graph in the y-axis. The outside term \(+2\) adds 2 to every y-value, so the graph then shifts 2 units upward. Therefore, option A is correct. Reflection in the x-axis would require \(-f(x)\), while a shift 2 units to the right would require \(f(x-2)\). Exam tip: changes inside the function affect the horizontal direction, whereas changes outside affect the vertical direction.

(|f(x)-2|) makes (f(x)-2) non-negative. Therefore the part with (f(x)<2) folds upward about the line (y=2).

In (|x-1|), inputs at equal distance from (x=1) become equal. Hence the graph is symmetric about (x=1).

(\frac{x^2-16}{x+4}=x-4), but (x=-4) is not in the domain. Therefore the line-like graph has a hole at (x=-4).

The direct answer is option A, \([0,1)\). The expression \(x-\lfloor x\rfloor\) is the fractional part of x. For every real x, the floor \(\lfloor x\rfloor\) is the greatest integer not exceeding x, so \(0\le x-\lfloor x\rfloor<1\). The value 0 occurs whenever x is an integer, for example x=2. The value 1 never occurs: just before the next integer, the fractional part gets close to 1 but remains less than 1. Option A is therefore correct. Option B, \((0,1]\), wrongly excludes 0 and includes the impossible value 1. Option C, \([0,1]\), wrongly includes 1. Option D, \(\mathbb R\), is wrong because the output is always between 0 and 1. Negative numbers cause no exception; for example, \(-1.3-\lfloor-1.3\rfloor=-1.3-(-2)=0.7\). Memory cue: fractional part includes 0 but never includes 1.