Graphs of standard functions

Correct answer: B. (4)

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.

Correct answer: C. \(5\) और \(-5\)

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 \(x=\pm a\).

Correct answer: A. 5 and −5

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 \(a>0\), \(x=\pm\sqrt{a}\).

Correct answer: B. (\frac{1}{4})

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.

Correct answer: C. (f(x)=x^3)

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

Correct answer: C. ( (0,-3) )

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

Correct answer: C. (y=\frac{1}{x})

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

Correct answer: B. (y=2x^2)

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

Correct answer: C. (4)

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.

Correct answer: C. (y=x^3)

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

Correct answer: B. x = -3 and x = 1

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.

Correct answer: A. ( [-5,\infty) )

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

Correct answer: A. ( {2} )

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

Correct answer: B. it is symmetric about the (y)-axis

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.

Correct answer: A. ( -1 )

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.

Correct answer: A. ( 2 )

(\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.

Correct answer: B. 1

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.

Correct answer: A. (2, 0)

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.

Correct answer: A. (-1,-1), (0,0), (1,1)

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.

Correct answer: C. 2

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.

Correct answer: A. x=0, x=1, and x=-1

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.

Correct answer: A. (2,0) and (6,0)

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.

Correct answer: A. x=-2, x=0, and x=2

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.

Correct answer: A. y=|x-1|

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.

Correct answer: A. (6,4)

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.