Graphs of standard functions

Correct answer: A. 2 units down

In \(f(x)=x^3-2\), the constant \(-2\) is applied outside the function, so every \(y\)-value decreases by 2. Therefore, the graph of \(y=x^3\) shifts 2 units downward. This is not a horizontal shift; horizontal changes occur inside the input, such as in \((x-2)^3\). Exam tip: a constant added or subtracted outside a function shifts its graph vertically up or down, respectively.

Correct answer: A. 2 इकाई बाईं ओर

A transformation inside the input of a function changes its horizontal position. The expression (x + 2)^3 is equal to (x - (-2))^3, so the graph of y = x^3 is translated 2 units to the left. The inflection point, originally (0, 0), moves to (-2, 0), which confirms the direction. Therefore option A is correct. The common mistake is to see the positive 2 and choose right, but inside changes use the opposite sign: x - h shifts right by h, while x + h shifts left by h. A vertical shift would require adding or subtracting a constant outside the cube, such as x^3 + 2 or x^3 - 2.

Correct answer: A. x-axis

The governing transformation rule is that replacing y=f(x) by y=−f(x) changes every output to its opposite while leaving the input coordinate unchanged. A point (a,a³) on y=x³ therefore becomes (a,−a³) on y=−x³. This is exactly reflection across the x-axis, because reflection in that axis preserves the x-coordinate and reverses the sign of the y-coordinate. For example, the point (2,8) changes to (2,−8), and the point (−1,−1) changes to (−1,1). Reflection in the y-axis would change (a,a³) to (−a,a³), which corresponds to replacing x by −x rather than placing a negative sign outside the function. Hence option A is correct; the two lines listed in options C and D are not the relevant axes of reflection.

Correct answer: A. ((-\infty,4)\cup(4,\infty))

(\frac{1}{x}) is never (0), so (\frac{1}{x}+4) is never (4). Thus all values except (4) are obtained.

Correct answer: A. First and third

In (\frac{2}{x}), (x) and (y) have the same sign. So the graph lies in the first and third quadrants.

Correct answer: B. Second and fourth

In (\frac{-3}{x}), (x) and (y) have opposite signs. So the graph lies in the second and fourth quadrants.

Correct answer: B. \(y=1\)

As \(|x|\) becomes very large, \(\frac{1}{x^2}\) approaches 0. Hence \(f(x)=\frac{1}{x^2}+1\) approaches 1, so its horizontal asymptote is \(y=1\). Option C, \(x=0\), is the vertical asymptote, not the horizontal one. Exam tip: find a horizontal asymptote by evaluating the function’s limit as \(x\to\pm\infty\).

Correct answer: C. -1

When \(x<-4\), we have \(x+4<0\). The signum function has value \(-1\) for a negative input, so \(f(x)=-1\). Remember that \(x=-4\) would give an input of zero and hence a value of \(0\), but equality is not included here.

Correct answer: A. It is discontinuous at every integer

At each integer \(n\), the left-hand limit is \(n-1\) while the right-hand limit is \(n\), producing a jump. Hence it is discontinuous at every integer. Exam tip: recognise this graph as a step graph.

Correct answer: B. -3

The floor function, written as ⌊x⌋, is defined as the greatest integer less than or equal to x. When x itself is an integer, that integer satisfies the condition and is already the greatest possible integer not exceeding x. Since -3 is an integer, ⌊-3⌋ = -3, so option B is correct. The result is not -4 because -4 is less than -3 and is therefore not the greatest eligible integer. It is not 3 because the floor function does not remove the negative sign, and it is not 0 because 0 is greater than -3 and does not satisfy the requirement of being less than or equal to the input. The distinction matters mainly for non-integer negative values, such as ⌊-3.2⌋ = -4.

Correct answer: A. (0)

(\lfloor 0.8\rfloor=0), because (0) is the greatest integer less than or equal to (0.8). For decimals, take the lower integer.

Correct answer: A. \(-1\)

The greatest integer function \(\lfloor x\rfloor\) gives the greatest integer less than or equal to \(x\). For every \(x\in[-1,0)\), this value is \(-1\), so the graph is the horizontal line \(y=-1\) on this interval. The value \(0\) begins at \(x=0\), which is excluded here. Exam tip: check the interval endpoints carefully when evaluating a floor function.

Correct answer: A. 0.6

The fractional part of a number \(x\) is defined as \(x-\lfloor x\rfloor\). Therefore, \(f(2.6)=\{2.6\}=2.6-\lfloor2.6\rfloor=2.6-2=0.6\). Hence, option A is correct. Exam tip: subtract the greatest integer less than or equal to the given number from the number itself.

Correct answer: B. 3

Every point on the y-axis has \(x=0\). Thus, \(f(0)=2(0)+3=3\), so the y-intercept is \(3\), and the intercept point is \((0,3)\). Remember that in \(y=mx+c\), the constant term \(c\) gives the y-intercept.

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

Every point on the \(x\)-axis has a \(y\)-coordinate of \(0\). Thus, set \(f(x)=0\): \(x^2-1=0\), so \(x^2=1\) and \(x=-1\) or \(x=1\). Therefore, the intercepts are \((-1,0)\) and \((1,0)\). Option B gives the point where \(x=0\), which lies on the \(y\)-axis instead. Exam tip: To find \(x\)-intercepts, set \(y=f(x)=0\).

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

The function can be written as \(f(x)=x^2+2x+1=(x+1)^2\). Comparing this with the standard form \(y=(x-h)^2+k\), we get \(h=-1\) and \(k=0\). Hence, the vertex is \((-1,0)\). The point \((0,1)\) is the y-intercept, not the vertex. Exam tip: After completing the square, read the vertex directly from \((x-h)^2+k\) as \((h,k)\).

Correct answer: A. 0

Completing the square gives f(x) = (x − 1)². A square is always nonnegative, so the minimum value is 0, attained at x = 1.

Correct answer: A. \((2,1)\)

The basic graph \(y=|x|\) has vertex \((0,0)\). In \(f(x)=|x-2|+1\), \(x-2\) shifts the graph 2 units to the right, while the outside \(+1\) shifts it 1 unit upward. Therefore, the vertex is \((2,1)\). Option D reverses the coordinate order. Exam tip: the vertex of \(y=|x-h|+k\) is \((h,k)\).

Correct answer: C. -2

The minimum value of \(|x+3|\) is 0, attained when \(x=-3\). Therefore, the minimum value of \(f(x)=|x+3|-2\) is \(0-2=-2\). Exam tip: the minimum value of \(|x-a|\) is 0, attained at \(x=a\).

Correct answer: B. 6

For every real \(x\), \(|x|\ge 0\), so \(-|x|\le 0\) and hence \(f(x)=-|x|+6\le 6\). At \(x=0\), \(f(0)=6\), so the maximum value is 6. The value 0 occurs at \(x=\pm6\), but it is not the maximum. Exam tip: the graph of \(-|x|\) opens downward, with its vertex at \(x=0\).

Correct answer: A. (0, 0)

Every point on the y-axis has \(x=0\). Thus, \(f(0)=0^3=0\), so the graph intersects the y-axis at \((0,0)\). Option D is incorrect because \(f(0)\) cannot be 1. Exam tip: To find a y-intercept, substitute \(x=0\).

Correct answer: B. \(x=-2\)

At an x-intercept, the function value is zero. Thus, \(x^3+8=0\), so \(x^3=-8\) and \(x=-2\). Therefore, the intercept is \((-2,0)\), whose x-coordinate is \(x=-2\). In such questions, set \(y=f(x)=0\); \(x=8\) is an incorrect choice based only on the constant term.

Correct answer: A. \((2,5)\)

The governing idea is the translation of the basic square-root graph \(y=\sqrt{x}\), whose endpoint is \((0,0)\). In \(f(x)=\sqrt{x-2}+5\), replacing \(x\) by \(x-2\) shifts the graph 2 units to the right, while adding 5 outside the radical shifts it 5 units upward. The endpoint therefore moves from \((0,0)\) to \((2,5)\). Equivalently, the radical begins when its radicand is zero: \(x-2=0\), so \(x=2\), and then \(f(2)=\sqrt{0}+5=5\). Thus option A is correct. Option B reverses the coordinates, option C uses the wrong horizontal sign, and option D ignores the vertical translation.

Correct answer: A. ((-1,-3))

The direct answer is A: (-1,-3). Start with the basic graph y=√x, whose starting point is (0,0). In y=√(x+1)-3, replacing x by x+1 shifts the graph one unit left, so the starting x-coordinate becomes -1. Subtracting 3 outside the root shifts every point three units down, so the y-coordinate becomes -3. Therefore the starting point is (-1,-3). Direct verification is also simple: the square root begins when its radicand is zero, so x+1=0 gives x=-1; then y=√0-3=-3. Option A is correct. Option B, (1,-3), would correspond to √(x-1)-3, a shift right. Option C, (-1,3), has the correct horizontal location but the wrong vertical shift; the graph moves down, not up. Option D, (0,-3), ignores the left shift caused by x+1. The domain begins at x=-1 and the range begins at y=-3. Memory cue: inside plus 1 means left 1, outside minus 3 means down 3.

Correct answer: B. (x=-5)

A vertical asymptote is a vertical line near which the function becomes unbounded because the function is not defined there. For a reciprocal function of the form one divided by an expression, first find where the denominator becomes zero. That value of x gives the possible vertical asymptote, provided no cancellation removes the restriction.

Here the denominator is x+5. Solving x+5=0 gives x=-5. Therefore the graph approaches the vertical line x=-5, so option B is correct. The lines y=5 and y=-5 are horizontal lines, not vertical ones, while x=5 would come from incorrectly changing the sign.