Graphs of standard functions

The direct answer is option A: the part below the x-axis is reflected above it. The expression \(|f(x)|\) changes every negative output into its positive opposite, while positive outputs remain unchanged. For example, if a point on \(y=f(x)\) is \((a,-3)\), then the new graph has \((a,3)\). A point with height 0 stays on the x-axis. Thus the x-coordinates do not change; only negative y-values turn upward. Option A is correct. Option B is wrong because the lower part does not disappear; it supplies the reflected upper part. Option C is wrong because reflection is vertical, not movement toward the y-axis. Option D is wrong because a negative value changes sign, so the graph does not remain unchanged wherever \(f(x)<0\). A useful rule is: \(y=|f(x)|\) folds everything below the x-axis upward. Memory cue: outside absolute value affects heights, not x-positions.

When (x<2), the larger of (x) and (2) is (2). For a maximum function, choose the larger expression on each interval.

The governing concept is the minimum operator together with the implication of the condition |x|≥2. Both |x| and 2 are non-negative, so squaring preserves the inequality and gives x²≥4. The function f(x)=min(x²,4) chooses the smaller of x² and 4. Since x² is at least 4 throughout the stated region, the smaller value is always 4. Equality occurs at |x|=2, and for larger absolute values x² is even greater, so the conclusion remains unchanged. Therefore option A is correct. Option B would apply where x²≤4, namely |x|≤2, while 2 and 0 are not selected by the minimum expression under the given condition.

When (x^2-1\le0), (|x^2-1|=1-x^2), so (-1\le x\le1). First determine the sign inside the modulus.

A vertical asymptote occurs at an excluded input value where the function becomes unbounded. For \(y=\frac{1}{x^2-4}\), the denominator must not be zero. Solve \(x^2-4=0\), giving \(x^2=4\) and therefore \(x=2\) or \(x=-2\). Near either value, the denominator approaches zero while the numerator remains 1, so the function grows without bound. Thus option A is correct.

The equations of vertical asymptotes are written using \(x\), because they describe vertical lines. The lines are \(x=-2\) and \(x=2\). Values such as \(y=-2\) and \(y=2\) would describe horizontal lines, not vertical asymptotes. Both denominator zeros must be included.

Since (|x+1|+2=|x-(-1)|+2), the vertex is ((-1,2)). In an upward-opening modulus graph, the vertex is the minimum point.

The graph of an inverse function is obtained by reflection in the line (y=x). In exams, check by changing a point ((a,b)) into ((b,a)).

It is the sum of distances of (x) from (2) and (6), so the minimum value is (6-2=4). For such questions, look at the distance between the two fixed points.

The governing concept is the end behavior of a rational function. In y=2/(x+1)−3, as x becomes very large positive or negative, the denominator x+1 also grows in magnitude, so 2/(x+1) approaches 0. Consequently y approaches 0−3=−3. The graph gets arbitrarily close to the horizontal line y=−3, which is its horizontal asymptote; option A is correct. The line x=−1 is the vertical asymptote because it makes the denominator zero. The values y=2 and x=3 do not describe the limiting behavior of this function. The numerator coefficient changes the approach pattern but not the horizontal asymptote.