Graphs of standard functions

In \(y=2|x|\), the output of the parent function \(|x|\) is multiplied by 2. Thus every \(y\)-value doubles while the \(x\)-values remain unchanged, producing a vertical stretch; the vertex stays at the origin. A horizontal stretch would involve changing the input, as in \(y=|2x|\). Exam tip: a multiplier outside the function causes a vertical transformation, whereas one inside the function affects the horizontal scale.

In \(y=|2x|\), the input \(x\) of the original function is replaced by \(2x\). Thus, for the same \(y\)-value, the required \(x\)-value is halved, so the graph moves toward the \(y\)-axis. This is a horizontal compression with scale factor \(1/2\). Option B is incorrect because a vertical change would result from multiplying the entire function by \(1/2\) outside the function. Exam tip: a multiplier inside the input causes a horizontal transformation, while a multiplier outside the function causes a vertical transformation.

In \(y=x^3-1\), 1 is subtracted from every value of \(y=x^3\). Thus each \(y\)-coordinate decreases by 1, so the entire graph shifts 1 unit downward. In contrast, \(y=(x-1)^3\) would shift the graph 1 unit to the right. Exam tip: a constant added or subtracted outside the function causes a vertical shift, whereas a change inside the input causes a horizontal shift.

The direct answer is option A: shift the graph 2 units to the right. Start with the basic graph y=x^3. For a horizontal translation, replacing x by x-h produces y=(x-h)^3 and moves the graph h units right. Here h=2, so y=(x-2)^3 is the graph of y=x^3 shifted 2 units right. For a point-based check, if the original graph has (0,0), after the shift it becomes (2,0), and indeed (2-2)^3=0. Option A is correct. Option B, 2 units left, would correspond to y=(x+2)^3. Option C, 2 units up, would correspond to y=x^3+2. Option D, 2 units down, would correspond to y=x^3-2. The change is inside the bracket, so it is horizontal, and x-2 means right rather than left. The supplied answer is correct.

At the intersection points, the two curves have equal \(y\)-values, so \(x^2=4\). Thus, \(x=-2\) and \(x=2\). The finite region lies between these two intersection points, giving \(-2\le x\le2\). Option B incorrectly excludes the negative \(x\)-values. Exam tip: To find the horizontal extent of a bounded region, first solve the equations of the two boundaries simultaneously.

For \(0\le x\le 4\), we have \(|x|=x\) and \(|x-4|=4-x\). Hence \(y=x+(4-x)=4\), a constant value, so the graph is horizontal throughout this interval. For \(x<0\) and \(x>4\), the slopes are \(-2\) and \(2\), respectively, so the graph is not horizontal there. Exam tip: For a function involving absolute values, first identify the points where the expressions inside the modulus signs become zero.

On (-3\le x\le 1), the sum ((x+3)+(1-x)=4) is constant. In such modulus sums the graph can be horizontal between the two points.

On swapping \(x\) and \(y\) in \(y=\frac{1}{x}\), we get \(x=\frac{1}{y}\), which rearranges to the same equation. Hence its graph is symmetric about \(y=x\); \(y=x^2\) is not. Exam tip: swap the variables to test this symmetry.

(f(-x)=\frac{-x}{x^2+1}=-f(x)). Therefore it is an odd function and its graph is symmetric about the origin.

The standard form of a modulus function is \(y=a|x-h|+k\), whose vertex is \((h,k)\). Here, \(x+2=x-(-2)\), so \(h=-2\) and \(k=5\). Therefore, the vertex is \((-2,5)\). Option B incorrectly changes the sign of the x-coordinate. Exam tip: Set the expression inside the modulus equal to zero to find the x-coordinate of the vertex.

For \(x\ge 0\), put \(|x|=x\), so \(x=3-x\) gives \(x=\frac{3}{2}\). Hence the intersection is \(\left(\frac{3}{2},\frac{3}{2}\right)\).

The minimum value of (|x-2|) is (0), so the maximum value of (y) is (3). An inverted modulus graph gives its maximum at the vertex.

The vertex occurs where the expression inside the absolute value is zero: \(x+4=0\), so \(x=-4\). Then \(f(-4)=|-4+4|-2=-2\), giving the vertex \((-4,-2)\).

Every point on the \(x\)-axis has a \(y\)-coordinate of 0. Thus, \(2|x+1|-6=0\), giving \(|x+1|=3\). Hence, \(x+1=3\) or \(x+1=-3\), so \(x=2\) or \(x=-4\). Therefore, the intersection points are \((-4,0)\) and \((2,0)\). The values in option B do not satisfy the equation. Exam tip: To find an \(x\)-intercept, set \(y=0\).

On (-5\le x\le 1), the sum is ((x+5)+(1-x)=6). In such graphs, the part between the two zero points is horizontal.

The modulus of any real number is non-negative, so \(y=|x^2-9|\) cannot be less than \(0\). This value is attained when \(x^2-9=0\), giving \(x=\pm3\). Hence, the correct answer is A. Exam tip: For a modulus function, first check whether the expression inside the modulus can equal zero; if it can, the minimum value is \(0\).

Completing the square gives \(y=x^2+4x+1=(x+2)^2-3\). In the standard form \(y=(x-h)^2+k\), the vertex is \((h,k)\); hence the vertex is \((-2,-3)\). Option B has the wrong sign for the \(x\)-coordinate. Exam tip: for \(x^2+bx+c\), the x-coordinate of the vertex is \(-\frac{b}{2}\).

On the x-axis, the y-coordinate is 0. Thus, \(x^2-8x+15=0\), which factors as \((x-3)(x-5)=0\). Therefore, \(x=3\) or \(x=5\), so option A is correct. Option B has incorrect signs for the roots. Exam tip: To find where a graph meets the x-axis, set \(y=0\).

In the equation \(x=-7\), the \(x\)-coordinate remains fixed at \(-7\), while \(y\) can be any real number. Therefore, its graph is a vertical line passing through \((-7,0)\), parallel to the \(y\)-axis. Option B would describe a horizontal line of the form \(y=\text{constant}\). Exam tip: remember that \(x=\text{constant}\) represents a vertical line, whereas \(y=\text{constant}\) represents a horizontal line.

For the function \(y=0\), the value of \(y\) is zero for every \(x\). Thus, all its points are of the form \((x,0)\), which lie on the x-axis. Therefore, its graph is the x-axis. Remember that \(x=0\) represents the y-axis, whereas \(y=0\) represents the x-axis.

For a square-root function, the radicand must satisfy \(x+2\geq 0\). The graph begins where the radicand is zero: \(x+2=0\), giving \(x=-2\). At this value, \(y=5-\sqrt{0}=5\), so the starting point is \((-2,5)\). Option B incorrectly uses \(x=2\), which is not the endpoint of the domain. Exam tip: a graph of \(\sqrt{x-a}\) starts at \((a,0)\), while an added constant shifts it vertically.

A vertical asymptote is a vertical line near which the graph moves without bound, while the function itself is not defined. For a rational function, it is usually found by making the denominator zero, provided no factor cancels with the numerator. This gives the excluded x-value and therefore the vertical line.

Here the denominator is \(x+5\). Setting it equal to zero gives \(x+5=0\), so \(x=-5\). Thus the graph has the vertical asymptote \(x=-5\), which is option A. The line \(y=0\) is horizontal, not vertical, and \(x=5\) would result from an incorrect sign.

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

The numerator and denominator have equal degree, so take the ratio of leading coefficients. Since (\frac{2}{1}=2), the asymptote is (y=2).

For this function, \(f(-x)=\frac{(-x)^2-1}{(-x)^2+1}=\frac{x^2-1}{x^2+1}=f(x)\). Hence, the function is even, so its graph is symmetric about the y-axis. Symmetry about the origin would require \(f(-x)=-f(x)\), which does not hold here. Exam tip: use \(f(-x)=f(x)\) for y-axis symmetry and \(f(-x)=-f(x)\) for origin symmetry.