Graphs of standard functions

The signum function \(\operatorname{sgn}(u)\) equals zero only when \(u=0\). Here, \(u=x+7\), so solving \(x+7=0\) gives \(x=-7\). Therefore, option A is correct. Exam tip: To find where \(\operatorname{sgn}(g(x))=0\), set the inner expression \(g(x)\) equal to zero.

Substituting \(x=5.9\) gives \(x-3=5.9-3=2.9\). The greatest-integer function \(\lfloor y\rfloor\) gives the greatest integer less than or equal to \(y\), so \(\lfloor 2.9\rfloor=2\). The value 2.9 is only the intermediate expression, not the final function value. Exam tip: for a decimal input, take the greatest integer that is less than or equal to it.

Substituting \(x=-0.4\) gives \(3x=3(-0.4)=-1.2\). The floor function \(\lfloor y\rfloor\) gives the greatest integer less than or equal to \(y\). Therefore, \(\lfloor -1.2\rfloor=-2\), so the correct answer is -2. Exam tip: for a negative decimal, the floor is the next lower integer, not the integer obtained by simply dropping the decimal part.

(\frac{7.4+2}{3}=\frac{9.4}{3}), which is greater than (3) and less than (4). So the greatest integer is (3).

The greatest integer function \(\lfloor x\rfloor\) gives the greatest integer less than or equal to \(x\). For every \(x\) in \([-3,-2)\), we have \(-3\le x<-2\), so \(\lfloor x\rfloor=-3\) and hence \(y=-3\). Note that \(x=-2\) is excluded from the interval; at \(x=-2\), the function value would be \(-2\).

Substituting \(x=1.35\) gives \(2x=2(1.35)=2.7\). Removing the integer part 2 leaves the fractional part \(0.7\), so \(f(1.35)=0.7\). Option B, \(2.7\), is the value of \(2x\) before taking its fractional part. In an exam, use \(\{y\}=y-\lfloor y\rfloor\) to evaluate such functions quickly.

The direct answer is option A, 6. A y-intercept is the point where a graph meets the y-axis. Every point on the y-axis has x=0, so substitute x=0 in 3x+2y=12: 3(0)+2y=12, hence 2y=12, and therefore y=6. Thus the intercept point is (0,6), while the question asks for its y-value, 6. Option A is correct because it gives 6. Option B, 4, would result from an incorrect calculation such as 12 divided by 3. Option C, 12, is the constant on the right side, not the y-intercept after solving the equation. Option D, 3, is the coefficient of x and is not obtained by putting x=0. A useful exam cue is: for a y-intercept, set x=0; for an x-intercept, set y=0. This reasoning is complete and the supplied answer is consistent.

Every point on the y-axis has \(x=0\). Thus, \(f(0)=0^3+3=3\), so the intercept point is \((0,3)\) and its y-coordinate is 3. Option B, 0, is the x-coordinate of the intercept, not its y-coordinate. Exam tip: To find a y-intercept, substitute \(x=0\).

The graph of \(y=x^2\) is an upward-opening parabola with vertex \((0,0)\), and its two sides are symmetric about the y-axis. Although \(y=|x|\) is also symmetric, it is V-shaped rather than parabolic. Exam tip: look for the quadratic power 2 to identify a parabola.

At every point on the \(x\)-axis, the \(y\)-coordinate is 0. Therefore, set \(f(x)=0\): \(x^3+27=0\), so \(x^3=-27\) and \(x=-3\). Hence, the intercept point is \((-3,0)\). Remember that an \(x\)-intercept is written as an ordered pair, not just as the \(x\)-value.

In \(f(x)=2x^3\), the y-value of \(y=x^3\) is multiplied by 2 for every fixed \(x\). Thus, the graph is vertically stretched by a factor of 2. It is not vertically compressed, since compression occurs when the multiplying coefficient lies between 0 and 1. Exam tip: In \(y=af(x)\), \(|a|>1\) causes a vertical stretch, while \(0<|a|<1\) causes a vertical compression.

The equation \(y^2=x+1\) can be rewritten as \(y=\pm\sqrt{x+1}\). Thus, for every \(x>-1\), there are two corresponding values of \(y\), so a vertical line can intersect the graph at two points. Therefore, it fails the vertical line test. In options B, C, and D, each allowed \(x\)-value gives exactly one \(y\)-value. Exam tip: a relation represents a function of \(x\) only if every vertical line intersects its graph at most once.

For (3) units right, use (x-3), and for (2) units down, use outside (-2). So the correct form is (y=(x-3)^2-2).

To shift the graph 4 units left, replace \(x\) by \(x+4\) inside the function. To shift it 5 units upward, add \(5\) outside the function. Thus, the required function is \(y=\sqrt{x+4}+5\). In option B, \(x-4\) shifts the graph 4 units right instead. Exam tip: a change inside the function acts in the opposite horizontal direction, while a change outside acts in the same vertical direction.

The transformation rule is that replacing \(x\) by \(x-h\) shifts a graph right by \(h\), while replacing it by \(x+h\) shifts it left by \(h\). Also, adding \(k\) outside the function shifts the graph vertically by \(k\). Starting with \(f(x)=1/x\), the expression \(1/(x+3)=f(x+3)\) shifts the graph 3 units to the left, because the inside sign is positive. The outside term \(-4\) then shifts the entire graph 4 units downward. Its vertical and horizontal asymptotes consequently move from \(x=0,y=0\) to \(x=-3,y=-4\). Therefore option A is correct; the other choices reverse one or both directions.

The central point of (y=x^3) is ((0,0)) and it shifts (4) right and (2) up. The negative sign changes direction but the central point remains ((4,2)).

As \(|x|\) becomes very large, \(\frac{4}{(x-3)^2}\) approaches 0. Hence, \(f(x)\) approaches 1, so the horizontal asymptote is \(y=1\). The line \(x=3\) is the vertical asymptote, not the horizontal one. Exam tip: For a function of the form \(\frac{a}{(x-h)^2}+k\), the horizontal asymptote is generally \(y=k\).

For \(f(x)=|x|\), we get \(f(-x)=|-x|=|x|=f(x)\). Hence, it is an even function and its graph is symmetric about the y-axis. The functions \(x\), \(x^3\), and \(\frac{1}{x}\) are odd functions, so their graphs are symmetric about the origin, not the y-axis. Exam tip: To test y-axis symmetry, check whether \(f(-x)=f(x)\).

The vertex of an absolute-value function occurs where the expression inside the modulus becomes zero. Thus, \(2x-8=0\) gives \(x=4\). Then \(f(4)=|0|+3=3\), so the vertex is \((4,3)\). Option C results from incorrectly taking \(x=8\), while the outside \(+3\) shifts the graph 3 units upward. Exam tip: Rewrite the function in the form \(a|x-h|+k\); its vertex is \((h,k)\).

The graph of \(f(x)=x\) is the line \(y=x\). When \(x>0\), \(y>0\), so the points lie in the first quadrant; when \(x<0\), \(y<0\), so the points lie in the third quadrant. Therefore, the correct answer is the first and third quadrants. Option B incorrectly assumes that \(x\) and \(y\) have opposite signs, which applies to a line such as \(y=-x\), not to \(y=x\). Exam tip: The graph of \(y=x\) passes through the origin and has equal-sign coordinates away from the origin.

Since (f(-x)=f(x)), the graph is symmetric about the (y)-axis. In exams, check (f(-x)) to identify an even function.

The direct answer is option B. A graph is symmetric about the origin when every point (x,y) is matched by the point (-x,-y). For f(x)=x^3, calculate f(-x)=(-x)^3=-x^3=-f(x). This is the definition of an odd function, and the graph of an odd function has origin symmetry. Option B states exactly this relation. Option A, f(-x)=f(x), defines an even function, whose graph is symmetric about the y-axis, not generally the origin. Option C, f(x)≥0, is false for negative x because x^3 is negative; it also does not state a symmetry rule. Option D, f(x)≤0, is false for positive x and likewise cannot establish origin symmetry. For example, (1,1) lies on the graph and its origin-reflected point (-1,-1) also lies on it. Remember: even means y-axis symmetry, odd means origin symmetry. The supplied answer is correct.

(|x|) is always (0) or greater and equals (0) at (x=0). In exams, remember the vertex of the modulus graph.

For real (\sqrt{x}), we need (x\ge 0). In exams, keep the expression inside a square root non-negative.

For \(f(x)=\frac{1}{x}\), the value at \(x=0\) is undefined, so the graph has no point on the line \(x=0\). Thus, \(x=0\) is the vertical asymptote and the graph does not intersect it. For comparison, with \(y=x\), the equation \(\frac{1}{x}=x\) gives \(x=\pm1\), so that line does intersect the graph. Exam tip: For \(y=\frac{1}{x}\), the asymptotes are \(x=0\) and \(y=0\).