Graphs of standard functions

(\lfloor -1.4\rfloor=-2), because (-2) is the greatest integer less than or equal to it. Be careful with negative decimals.

For every (x\in[2,3)), (\lfloor x\rfloor=2). So this part of the graph is the horizontal line (y=2).

The greatest integer function \\(f(x)=\lfloor x\rfloor\\) remains constant at the value \\(n\\) on every interval \\([n,n+1)\\), so its graph consists of horizontal steps. Each step includes its left endpoint and excludes its right endpoint. A parabola or a circle does not represent these constant intervals. Exam tip: \\(\lfloor x\rfloor\\) means the greatest integer less than or equal to \\(x\\).

The graph \(y=x^2\) is an upward-opening parabola. Its minimum \(y\)-value is \(0\), attained at \(x=0\), so its vertex is \((0,0)\). Exam tip: in \(y=(x-h)^2+k\), the vertex is \((h,k)\).

The function is in vertex form \(y=(x-h)^2+k\), where \(h=-2\) and \(k=0\). Thus, its graph is \(y=x^2\) shifted 2 units to the left, giving the vertex \((-2,0)\). Option A has the wrong sign for the x-coordinate. Exam tip: the vertex of \((x+a)^2\) is \((-a,0)\).

For every real \(x\), \(x^2\ge 0\). Therefore, \(f(x)=x^2-3\ge -3\). This minimum is attained at \(x=0\), since \(f(0)=0^2-3=-3\). The value 0 is the minimum of \(x^2\) alone, not of the entire function. Exam tip: For a function containing \(x^2\), first use the fact that its minimum value is 0.

For every real \(x\), \((x-2)^2\ge 0\). Hence \(f(x)=-(x-2)^2\le 0\). At \(x=2\), \(f(2)=0\), so the maximum value is 0. Exam tip: a quadratic with a negative coefficient of the squared term opens downward, so its vertex gives the maximum value.

The absolute value is smallest when its expression is zero: \(x+1=0\), so \(x=-1\) and \(f(-1)=0\). Thus, the vertex is \((-1,0)\).

The minimum value of (|x|) is (0), so the minimum value of (|x|-2) is (-2). The graph shifts (2) units down.

The basic graph y = |x| has vertex (0, 0). Replacing x by x - 3 translates that graph 3 units to the right, so the vertex moves to (3, 0). The negative sign outside the absolute value reflects the graph across the x-axis, changing the V-shape into an inverted V, but it does not move the point where the absolute-value expression is zero. Algebraically, |x - 3| = 0 when x = 3, and then f(3) = -0 = 0. Therefore the vertex is (3, 0), making option A correct. The other choices confuse the horizontal shift with a sign change or interchange the coordinates.

A point \((x,y)\) lies on the graph only when \(y=|x|\). The points \((2,2)\), \((-2,2)\), and \((0,0)\) satisfy this condition. For \((-3,-3)\), however, \(|-3|=3\), not \(-3\), so the point does not lie on the graph. Exam tip: In \(y=|x|\), the value of \(y\) is always zero or positive.

For a function graph, \(y=f(x)\). Substituting \(x=2\) gives \(y=2^3=2\times2\times2=8\), so the correct answer is 8. Values such as 6 or 9 may result from incorrectly calculating the cube. Exam tip: \(x^3\) means multiplying \(x\) by itself three times, not multiplying it by 3.

Since \(f(-x)=(-x)^3=-x^3=-f(x)\), the function \(f(x)=x^3\) is odd. The graph of an odd function is symmetric about the origin, so option B is correct. Symmetry about the y-axis would require an even function satisfying \(f(-x)=f(x)\). Exam tip: test \(f(-x)=-f(x)\) to identify origin symmetry.

Adding 1 outside the expression changes every output value by 1, so \(y=x^3+1\) is obtained by moving each point of \(y=x^3\) one unit upward. Thus the graph shifts 1 unit up, not horizontally. Exam tip: a constant added outside a function causes a vertical shift, whereas \(g(x-a)\) shifts a graph to the right.

In ((x-1)^3), the inside term is (x-1), so the graph shifts (1) unit right. An inside change shows the opposite direction.

The graph begins where the radicand first becomes zero: \(x-1=0\), so \(x=1\). Then \(f(1)=0\), giving the starting point \((1,0)\).

A point \((x,y)\) lies on the graph when \(y=\sqrt{x}\). For \((4,2)\), \(\sqrt{4}=2\), so it satisfies the equation. In \((2,4)\), \(\sqrt{2}\neq4\); \((-4,2)\) is invalid because the square-root function is not defined for a negative real input; and \((4,-2)\) is impossible because the principal square root is non-negative. Exam tip: substitute the point’s \(x\)-coordinate into the function and compare the result with \(y\).

For a reciprocal function, a vertical asymptote occurs where the denominator becomes zero, because the function is undefined there and its values grow without bound near that x-coordinate. Set the denominator equal to zero: x + 1 = 0, which gives x = -1. Thus the vertical asymptote is the vertical line x = -1, so option B is correct. The equation y = 1 is not an asymptote of this translated reciprocal graph; it is a horizontal line that the graph does not approach as x becomes large. The lines x = 1 and y = -1 result from incorrect signs or from confusing vertical and horizontal asymptotes.

The value of (\frac{1}{x}) approaches (0) for large (x). So the horizontal asymptote of (\frac{1}{x}+2) is (y=2).

The circle (x^2+y^2=1) gives two (y)-values for some (x)-values. So it fails the vertical line test.

A function must have only one (y) for each (x). Therefore a vertical line cuts it at at most one point.

\(f(2)=2^2=4\) and \(f(-2)=(-2)^2=4\), so the two values are equal. The square of a negative number is also positive. Option D is incorrect because \(f(-2)\) is \(4\), not \(-4\). Exam tip: When \(f(-x)=f(x)\), the function is even and its graph is symmetric about the \(y\)-axis.

\(f(2)=2^3=8\) and \(f(-2)=(-2)^3=-8\), so \(f(-2)=-f(2)\). This shows that \(f(x)=x^3\) is an odd function, whose graph is symmetric about the origin. Exam tip: If \(f(-x)=-f(x)\), the function is odd.

A point on the x-axis has y-coordinate 0. Thus, \(x^2=0\), giving \(x=0\) and the point \((0,0)\). The parabola \(y=x^2\) has its vertex at \((0,0)\), where it touches the x-axis. The points \((1,0)\) and \((-1,0)\) are incorrect because the function values there are 1, not 0. Exam tip: Since \(x^2\) is always non-negative, the graph does not cross the x-axis.

For \(f(x)=|x|\), the value is \(|x|\ge 0\). When \(x\ge0\), \(|x|=x\) (a line with slope \(1\)); when \(x<0\), \(|x|=-x\) (a line with slope \(-1\)). These two rays meet at \(x=0\), giving a V-shape with vertex at \((0,0)\). The function depends only on \(x\) through \(|x|\), so it is symmetric about the y-axis.