Graphs of standard functions

The direct answer is option B: 4. The function rule is y=f(x)=\sqrt{x}. To find y when x=16, substitute 16 for x: y=\sqrt{16}=4 because 4×4=16. Thus the graph contains the point (16,4). Option A, 2, is wrong because 2×2=4, not 16. Option B, 4, is correct because its square is 16. Option C, 8, is wrong because 8×8=64. Option D, 16, is wrong because \sqrt{16} is 4, not 16; confusing x with y is a common mistake. Also, the square-root function has domain x≥0, and 16 is allowed. Its output is the nonnegative principal square root, so we choose 4 rather than −4. A useful exam cue is to ask, “Which nonnegative number multiplied by itself gives x?” For 16, that number is 4.

For the given graph, setting \(y=5\) gives \(|x|=5\). A number with absolute value 5 can be either \(5\) or \(-5\), so option C is correct. Choosing only \(5\) ignores the negative value, which is the error in option A. Exam tip: for \(|x|=a\) with \(a>0\), write \(x=\pm a\).

Substituting \(y=25\) in the function gives \(x^2=25\). Hence, \(x=5\) or \(x=-5\), so both values are possible. Choosing only 5 or only −5 is incorrect because both a positive and a negative number can have the square 25. In exams, remember that for \(x^2=a\) with \(a>0\), \(x=\pm\sqrt{a}\).

The direct answer is B: 1/4. The function is f(x)=1/x, and on its graph y=f(x). Substitute x=4: y=1/4. Therefore the point is (4,1/4). Option A, 4, wrongly treats y as equal to x. Option B, 1/4, follows directly from the reciprocal rule. Option C, −4, has the wrong sign and magnitude. Option D, −1/4, has a wrong negative sign; a positive x gives a positive reciprocal. Also, x=0 is not allowed, but x=4 is allowed. Memory cue: for y=1/x, put the given x in the denominator and keep the sign; positive input gives positive output, negative input gives negative output.

(x^3) can take every real (y)-value. In exams, take the range of the cubic function as all real numbers.

The vertex of (y=|x|) is ((0,0)), and (-3) shifts it downward. In exams, add vertical shifts directly to the (y)-coordinate.

The graph (y=\frac{1}{x}) approaches the axes as asymptotes. In exams, remember both asymptotes of the reciprocal graph.

For vertical stretching, multiply the whole function by (2). In exams, treat an outside multiplier as vertical scaling.

Because the coefficient is negative, the parabola opens downward and the vertex (y)-value (4) is maximum. In exams, the vertex is maximum for a downward parabola.

(y=x^3) is negative for negative (x) and positive for positive (x). In exams, remember the cubic graph crosses the (x)-axis.

The governing concept is finding x-intercepts by setting the output y equal to zero. For the graph y = |x + 1| - 2, put 0 = |x + 1| - 2, which gives |x + 1| = 2. A modulus equation |u| = 2 has two cases: u = 2 or u = -2. Therefore x + 1 = 2 gives x = 1, and x + 1 = -2 gives x = -3. Both values satisfy the original equation, so the graph cuts the x-axis at x = -3 and x = 1. Hence option B is correct. The other choices arise from ignoring the negative modulus case, changing the sign incorrectly, or shifting the constants improperly.

For (\sqrt{x+5}), (x+5\ge 0), so (x\ge -5). In exams, keep the expression inside the square root at least (0).

In a constant function, (y) is always (2). In exams, the range of a constant line is the set containing only that (y)-value.

The domain of (y=x^2) is all real numbers and its graph is symmetric about the (y)-axis. In exams, remember the domain and range of a parabola separately.

In the signum function, the value is (-1) for negative (x). In exams, keep the three cases (x<0), (x=0), and (x>0) separate.

(\lfloor 2.7\rfloor=2) because it is the greatest integer less than or equal to (2.7). In exams, do not round; move down to the greatest integer.

The governing concept is the floor function, defined by ⌊x⌋ equal to the greatest integer less than or equal to x. For every x in the half-open interval [1, 2), the inequalities are 1 ≤ x < 2. The greatest integer not exceeding any such x is therefore 1, so f(x) = ⌊x⌋ = 1 throughout the interval. Thus option B is correct. The left endpoint x = 1 is included and gives 1, whereas x = 2 is excluded because the value would jump to 2 there. This explains the half-open notation. Option D is incorrect because the floor graph is stepwise rather than the continuous line y = x; options A and C do not match the defining integer value on the interval.

Answer: option A, (2,0). Every point on the x-axis has y-coordinate zero. Therefore, to find where the graph meets the x-axis, put y=0 in y=x³−8. We obtain 0=x³−8, so x³=8. The real cube root of 8 is 2, giving the point (2,0). Substitution verifies it: 2³−8=8−8=0. Option B, (−2,0), is wrong because (−2)³−8=−8−8=−16, not zero. Option C, (0,8), is the y-intercept, obtained by setting x=0. Option D incorrectly uses the constant 8 as the x-coordinate without solving the equation. Remember: x-axis means y=0, while y-axis means x=0.

Two function graphs intersect where their y-values are equal for the same x-value. Therefore set x^3=x and solve: x^3−x=0. Factoring gives x(x^2−1)=0, and then x(x−1)(x+1)=0. Hence the possible x-values are −1, 0, and 1. Since the second graph is y=x, the corresponding y-values are exactly −1, 0, and 1. The intersection points are consequently (−1,−1), (0,0), and (1,1), making option A correct. Option B omits the valid negative solution. Option C assigns the wrong signs to the nonzero y-coordinates. Option D recognizes only the central intersection and misses the two additional points.

The governing concept is piecewise evaluation of absolute values using the sign of each expression. For x≥2, both x and x−2 are nonnegative. Therefore |x|=x and |x−2|=x−2. Substitute these forms into the function: y=x−(x−2)=x−x+2=2. The variable terms cancel for every x in the stated interval, so the graph is horizontal there and has the constant value 2. Thus option C is correct. Option A would follow from assigning an incorrect negative sign to the second absolute value. Option B loses the remaining constant term. Option D stops before simplification and incorrectly leaves a variable expression, even though the x terms cancel completely.

The governing concept is the condition for intersection: at a common point, the two functions must have equal y-values for the same x. Hence solve x^2=|x|. Since x^2=|x|^2 for every real x, this becomes |x|^2=|x|, or |x|(|x|−1)=0. Thus |x|=0 or |x|=1. The first condition gives x=0; the second gives x=1 and x=−1. Therefore the complete set is x=−1, 0, and 1, so option A is correct. Option B omits x=±1, and option C omits x=0 and x=−1. For x=±2, x^2=4 while |x|=2, so option D does not represent intersections.

An x-axis intercept is a point whose y-coordinate is zero. Set the quadratic equal to zero: x^2−8x+12=0. Factorisation gives (x−2)(x−6)=0. By the zero-product property, x−2=0 or x−6=0, so x=2 or x=6. The corresponding points on the x-axis are (2,0) and (6,0). Thus option A is correct. Option B changes the signs of both roots and does not satisfy the equation. Option C lists points with x=0, which concerns the y-axis rather than the x-axis. Option D gives 3 and 4, but substituting either value does not make the quadratic zero. The factorisation also provides a direct verification of both intercepts.

At an intersection, both functions have the same y-value for the same x, so solve x^3=4x. Move all terms to one side: x^3−4x=0. Factorising gives x(x^2−4)=0, and the difference of squares gives x(x−2)(x+2)=0. Therefore the three real solutions are x=−2, x=0, and x=2. Substitution confirms them: at x=2, both functions equal 8; at x=−2, both equal −8; and at x=0, both equal 0. Hence option A is correct. Option B corresponds to an incorrect factor x^2−1. Option C treats the coefficient 4 as a root, and option D ignores the two nonzero solutions.

The governing concept is the principal-square-root identity √(a²)=|a| for every real number a. The square root symbol denotes the non-negative root, so √(a²) cannot generally be replaced by a without considering its sign. Taking a=x−1 gives f(x)=√((x−1)²)=|x−1|. Consequently, the graph is the usual V-shaped graph y=|x| translated one unit to the right, with vertex at (1,0), so option A is correct. Option B gives only a straight line and is negative for x<1. Option C is the original parabola before the square root is taken, and option D has a different curve and domain restriction x≥1. The given function is defined for every real x.

The governing transformation rule for a parabola in vertex form is y=(x-h)^2+k, whose vertex is (h,k). Comparing y=(x-6)^2+4 with this form gives h=6 and k=4, so the vertex is (6,4). Equivalently, replacing x by x-6 shifts y=x^2 six units to the right, while adding 4 shifts it four units upward. Hence option A is correct. Option B has the horizontal direction reversed, option C reverses the vertical shift, and option D interchanges the x- and y-coordinates. The original parabola has vertex (0,0), which confirms these two translations.