Graphs of standard functions

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

A graph is symmetric about the \(y\)-axis if and only if \(f(-x)=f(x)\), which means the function is even. For option A, \(f(-x)=(-x)^2+1=x^2+1=f(x)\), so it is correct. Options B and C do not satisfy this condition, and \(\sqrt{x}\) is defined only for \(x\geq 0\). Exam tip: Test \(y\)-axis symmetry by calculating \(f(-x)\).

A graph is symmetric about the origin when its function is odd, meaning \(f(-x)=-f(x)\). For option A, \(f(-x)=(-x)^3=-x^3=-f(x)\), so the graph of \(y=x^3\) is origin-symmetric. In contrast, \(x^2\) and \(|x|\) are even functions, whose graphs are symmetric about the y-axis. Exam tip: test \(f(-x)=-f(x)\) for origin symmetry.

The vertical line test determines whether a curve represents a function of \(x\): every vertical line must intersect the graph at no more than one point. For the circle \(x^2+y^2=4\), solving for \(y\) gives \(y=\pm\sqrt{4-x^2}\). For example, when \(x=0\), the graph contains both \((0,2)\) and \((0,-2)\), so the vertical line \(x=0\) intersects it twice. Hence the circle fails the test and is not the graph of a single-valued function \(y=f(x)\). The parabolas or absolute-value and square-root graphs in the other options give at most one \(y\)-value for each permitted \(x\), so option C is correct.

For a real square root, (9-x\ge 0) is required, so (x\le 9). In square root graphs, keep the inside expression (0) or positive.

To find where a graph cuts the x-axis, use the defining condition y=0. Substituting into y=x^3+27 gives x^3+27=0, so x^3=−27. The real cube root is x=−3, and therefore the x-axis intercept is (−3,0), making option A correct. Substitution confirms the result because (−3)^3+27=−27+27=0. Option B results from dropping the negative sign while taking the cube root. Option C is the y-intercept, since x=0 gives y=27 rather than zero. Option D incorrectly uses the constant 27 as the x-coordinate; substituting x=27 gives a positive value, not an x-axis point. The negative sign is essential in this calculation.

The governing concept is the definition of a y-intercept: it is the point where a graph crosses the y-axis, and every point on that axis has x=0. Substitute x=0 into y=2|x|−3. This gives y=2|0|−3=0−3=−3, so the y-intercept is (0,−3), making option A correct. Option B has the wrong sign. Options C and D have y=0, so they are written as x-axis points rather than y-axis points. In fact, solving 2|x|−3=0 gives |x|=3 and the x-intercepts (3,0) and (−3,0). This comparison clarifies why the requested intercept must be found by setting x, not y, equal to zero.

The expression is in vertex form y=a(x−h)²+k. Since a is negative, the parabola opens downward, so its vertex gives the maximum value. More directly, (x−1)² is always nonnegative. Multiplying it by −3 produces a quantity that is always less than or equal to zero. Consequently, y=−3(x−1)²+4 is always less than or equal to 4. Equality occurs when (x−1)²=0, that is, when x=1. Substitution gives y=−3(0)+4=4. Therefore the maximum value is 4 and option A is correct. The number 1 is the x-coordinate of the vertex, not the maximum y-value; −3 is the coefficient of the squared term, and −4 is not obtained from the expression. The negative leading coefficient confirms that the vertex represents a maximum rather than a minimum.

The ceiling function ⌈x⌉ is defined as the least integer greater than or equal to x. Because 4.1 lies strictly between 4 and 5, the smallest integer that is at least 4.1 is 5. Hence g(4.1) = ⌈4.1⌉ = 5, so option A is correct. A common error is to use the floor function: ⌊4.1⌋ = 4, which explains option B but does not follow the ceiling definition. Option C is simply the input and is not an integer, whereas a ceiling value must be an integer. Option D has no connection with the stated rule. The result illustrates the step-like behavior of the ceiling function for non-integer inputs.

For a rational function, the vertical asymptote is found by setting the denominator equal to zero after checking that no factor cancels with the numerator. In y = 1/(x − 6), the denominator is zero when x − 6 = 0, which gives x = 6. Near this x-value, the denominator becomes very small, so the function grows without bound in magnitude; this is the behavior that defines the vertical asymptote. The numerator is 1 and cannot cancel anything. Option B comes from changing the sign incorrectly. Options C and D are horizontal lines; in fact y = 0 is the horizontal asymptote of this graph, while y = 6 is not an asymptote. Therefore option A is the unique correct answer.

\(y=x^3\) is an odd function because \(f(-x)=(-x)^3=-x^3=-f(x)\). Hence, its graph is symmetric about the origin. In contrast, \(y=x^2\) is symmetric about the y-axis. Exam tip: test \(f(-x)=-f(x)\) to identify odd functions.