Graphs of standard functions

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.

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.

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.

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

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

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

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 \(x\to\pm\infty\).

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.

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.

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.

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

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.

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.

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.

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 \(y=f(x)=0\).

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 \((h,k)\).

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

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 \(x=a\).

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 \(x=0\).

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 \(x=0\).

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.

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.

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.

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.