Graphs of standard functions

The graph of a function consists of ordered pairs \((x,f(x))\). Here, \(f(1)=4\), \(f(2)=3\), \(f(3)=2\), and \(f(4)=1\), so the graph is \(\{(1,4),(2,3),(3,2),(4,1)\}\). Option B represents the identity function \(f(x)=x\), while option D is incomplete because it omits \((1,4)\). Exam tip: the graph of a function must contain exactly one ordered pair for every element of its domain.

The minimum of (|x-2|) is (0) at (x=2), so (y=3). In exams, set the inside of the modulus equal to (0).

The parabola opens downward and its vertex is ((-4,5)). In exams, identify the vertex ((h,k)) in (y=a(x-h)^2+k).

For the square root, (7-2x\ge 0), so (x\le \frac{7}{2}). In exams, keep the expression inside the square root non-negative.

The denominator (x-1) gives vertical asymptote (x=1), and the outside (-2) gives horizontal asymptote (y=-2). In exams, check both the denominator and vertical shift.

For (x\ge 1), (|x+3|=x+3) and (|x-1|=x-1), so (y=4). In exams, open modulus expressions interval-wise.

For every (x\in[-2,-1)), the greatest integer is (-2). In exams, for negative decimals choose the lower integer.

The governing concept is the piecewise definition of the signum function applied to a shifted input. For any real t, sgn(t) = -1 when t < 0, 0 when t = 0, and 1 when t > 0. At the specified point x = 5, the inner expression is x - 5 = 5 - 5 = 0. Therefore f(5) = sgn(0) = 0, making option B correct. The value -1 occurs on the branch x < 5, while 1 occurs on the branch x > 5; these describe the left and right horizontal portions of the graph, not the value at the jump point. The number 5 is the input coordinate, not the output. The graph includes the point (5, 0).

The direct answer is A: (2,0) and (4,0). Points on the x-axis have y=0, so solve x²−6x+8=0. Factor the quadratic: x²−6x+8=(x−2)(x−4). Thus x=2 or x=4. With y=0, the intercepts are (2,0) and (4,0). Option A is correct. Option B uses negative roots, but substituting −2 or −4 does not make the expression zero. Option C gives points on the y-axis because their x-coordinate is 0; these are not x-intercepts. Option D uses 1 and 8, which are not the roots of the quadratic. A quick check is f(2)=4−12+8=0 and f(4)=16−24+8=0. Memory cue: for x-axis intersections, set y or f(x) equal to zero, then write coordinates with y=0.

A point on the \(x\)-axis has \(y=0\). Hence, \(2|x|-6=0\), giving \(|x|=3\), so \(x=-3\) or \(x=3\). Therefore, option A is correct. In option C, \(x=0\) gives the vertex \((0,-6)\), which is not on the \(x\)-axis. Exam tip: when \(|x|=a\) with \(a>0\), consider both values \(x=\pm a\).

Since (\sqrt{x+2}\ge0), (f(x)\ge -4). In exams, take the square-root minimum (0) and add the vertical shift.

The governing graph-transformation rule is that replacing x by x−h in y=f(x) shifts the graph h units horizontally to the right, while adding k outside the function shifts it k units vertically upward. Starting with the parent cubic y=x^3, the expression (x−2)^3 therefore moves every point 2 units to the right. The final +1 then raises every resulting point by 1 unit, giving y=(x−2)^3+1. Hence option B is correct. Option A reverses both directions. Option C names the correct distances but interchanges horizontal and vertical movement. Option D gives the opposite shifts. The order of the two translations does not affect the final result, because both are applied to the same graph.

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

The denominator (x+3) becomes zero at (x=-3). In exams, find the vertical asymptote of a reciprocal graph from the denominator.

From (2x+6=0), (x=-3) and the minimum is (0). In exams, find the vertex by setting the inside of the modulus to (0).

When (x\in[5,6)), (x-2\in[3,4)), so (\lfloor x-2\rfloor=3). In exams, first transform the inside interval.

Squaring gives (x^2+y^2=9) with (y\ge0). In exams, identify (\sqrt{r^2-x^2}) as the upper semicircle.

When \\(x<0\\), we have \\(|x|=-x\\). Hence, \\(f(x)=|x|-x=-x-x=-2x\\), so the graph coincides with the part of the line \\(y=-2x\\) for which \\(x<0\\). Exam tip: resolve the modulus according to the sign of the variable before simplifying.

(\frac{x}{|x|}=1) for (x>0) and (-1) for (x<0). In exams, remember to exclude (x=0) from the domain.

Since (\sqrt{x-1}\ge0) and its least value is (0) at (x=1), (y\le3). In exams, for a negative square-root graph, check the starting point for maximum.

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

The degrees of numerator and denominator are equal, so the ratio of leading coefficients is (\frac{2}{1}=2). In exams, use the ratio of leading coefficients for equal degrees.

(\lfloor3.1\rfloor+1=4), so ((3.1,3)) is not on the graph. In exams, first find the greatest integer of (x) while checking a point.

The distance between (4) and (-2) is (6), so the minimum sum is (6). In exams, remember the minimum of (|x-a|+|x-b|) is (|a-b|).

The minimum of (|x+1|) is (0) at (x=-1), so (y=6) is maximum. In exams, a modulus graph with a negative multiplier has its maximum at the vertex.