Graphs of standard functions

For (2) units left, use (x+2), and for (3) units up, use outside (+3). Thus the correct form is (y=(x+2)^2+3).

For (4) units right, use (x-4), and for (1) unit down, use outside (-1). So (y=|x-4|-1) is correct.

Since \(g(x)=f(x)+3\), the function value increases by 3 for every \(x\). Therefore, each point on the graph of \(f\) moves 3 units upward to form the graph of \(g\). Remember for exams: \(y=f(x)+k\) shifts a graph \(k\) units upward, whereas \(y=f(x)-k\) shifts it downward.

The direct answer is A: the graph moves 2 units to the right. Begin with the standard graph \(y=1/x\). Replacing x by \(x-2\) gives \(y=1/(x-2)\), which is a horizontal translation right by 2 units. This is also confirmed by the vertical asymptote: the denominator is zero when \(x-2=0\), so \(x=2\), whereas the original asymptote is x=0. Option A is correct. Option B would correspond to \(1/(x+2)\), a left shift. Option C would require adding 2 outside the fraction, such as \(1/x+2\), which is a vertical shift up. Option D would require subtracting 2 outside, not changing x inside the denominator. Memory cue: inside x, the sign appears reversed: x−2 means right 2 and x+2 means left 2.

(\frac{5.8-1}{2}=2.4) and (\lfloor2.4\rfloor=2). First find the inside value and then apply the greatest integer function.

At every point on the x-axis, \(y=f(x)=0\). Thus, \(-(x-1)^2+4=0\), which gives \((x-1)^2=4\). Therefore, \(x-1=\pm2\), so \(x=-1\) or \(x=3\). Hence, option A is correct. Remember that the complete intercept points are \((-1,0)\) and \((3,0)\); this question asks only for their x-coordinates.

The graph of y = |x| has vertex (0, 0). Replacing x by x − 5 shifts it 5 units right, and subtracting 2 shifts it 2 units down. Therefore, the vertex is (5, −2).

The modulus of any real number is non-negative, so \(|2x+4|\geq 0\). It attains its minimum value when \(2x+4=0\), giving \(x=-2\). Therefore, \(f(x)=0+1=1\). Hence, option A is correct; 0 is only the minimum value of the modulus term, not of the whole function. Exam tip: For an expression of the form \(|g(x)|+c\), check whether \(g(x)=0\) has a real solution; if it does, the minimum is \(c\).

For a modulus function \(f(x)=|ax+b|\), the vertex occurs where the expression inside the modulus is zero. Thus, \(3x-9=0\), giving \(x=3\), and the vertex is \((3,0)\). The choice \(x=9\) results from confusing the constant term with the required value of \(x\). Exam tip: Set the linear expression inside the modulus equal to zero to find the vertex's \(x\)-coordinate.

At an \(x\)-intercept, the function value is zero. Thus, \(2|x+2|-6=0\), giving \(|x+2|=3\). Therefore, \(x+2=3\) or \(x+2=-3\), so \(x=1\) or \(x=-5\). Hence, the intercepts are \((1,0)\) and \((-5,0)\). For an \(x\)-intercept, remember to set \(y=f(x)=0\).

Since \(|x-1|\ge 0\), we have \(-|x-1|\le 0\), so \(f(x)=-|x-1|+4\le 4\). At \(x=1\), \(|x-1|=0\) and \(f(1)=4\); hence the maximum value is 4. Option 1 may result from confusing the \(x\)-value at which the maximum occurs with the function's maximum value. Exam tip: a negative coefficient of the modulus gives an inverted V-shaped graph, whose vertex represents the maximum.

For an x-intercept, set \(f(x)=0\): \(3(x-1)^2-12=0\). Thus \((x-1)^2=4\), so \(x-1=\pm2\), giving \(x=-1\) and \(x=3\). Therefore, the intercept points are \((-1,0)\) and \((3,0)\). Option B includes \(x=1\), which is the x-coordinate of the vertex, not an x-intercept. Exam tip: To find x-intercepts, always set \(y=f(x)=0\).

For a quadratic function \(f(x)=ax^2+bx+c\), the axis of symmetry is \(x=-\frac{b}{2a}\). Here, \(a=1\) and \(b=-10\), so \(x=-\frac{-10}{2(1)}=5\). Therefore, the correct axis is \(x=5\). The option \(x=-5\) results from a sign error, while \(y=5\) is not the correct form of the axis for this graph. Exam tip: Identify the coefficients of \(x^2\) and \(x\), then apply \(-\frac{b}{2a}\) directly.

For a quadratic function \(f(x)=ax^2+bx+c\), the axis of symmetry is \(x=-\frac{b}{2a}\). Here, \(a=2\) and \(b=-8\), so \(x=-\frac{-8}{2\cdot2}=2\). Therefore, the correct answer is \(x=2\). Remember that the axis of symmetry is a vertical line, so \(y=2\) cannot be correct here.

Completing the square gives \(f(x)=x^2+2x+5=(x+1)^2+4\). Since \((x+1)^2\geq 0\), its least value is 0, attained at \(x=-1\). Therefore, the minimum value of the function is 4. Exam tip: For a quadratic written as \((x-h)^2+k\), the minimum value is directly k.

Completing the square gives \(f(x)=-(x-3)^2+4\). Since \((x-3)^2\geq 0\), we have \(-(x-3)^2\leq 0\), so the maximum value of the function is 4, attained at \(x=3\). Exam tip: For a downward-opening quadratic, the vertex gives the maximum value.

The direct answer is A, \((-6,-2)\). The basic square-root graph \(y=\sqrt{x}\) starts at \((0,0)\), because its input cannot be negative and \(\sqrt0=0\). In \(y=\sqrt{x+6}-2\), require \(x+6\ge0\), so the first possible x-value is \(-6\). At x=−6, \(y=\sqrt0-2=-2\). Therefore the starting point is \((-6,-2)\). Option A is correct. Option B, \((6,-2)\), uses the wrong horizontal sign; x+6 shifts left, not right. Option C, \((-6,2)\), has the vertical sign wrong because the outside −2 moves the graph down. Option D, \((0,-2)\), ignores the horizontal shift and is not the endpoint. Remember: \(\sqrt{x-h}+k\) starts at \((h,k)\).

A vertical asymptote occurs where the denominator becomes zero. Here, \(x-5=0\), so the vertical asymptote is \(x=5\). The line \(y=2\) is the horizontal asymptote, not the vertical one. Exam tip: for \(\frac{1}{x-a}+b\), the vertical asymptote is \(x=a\) and the horizontal asymptote is \(y=b\).

As \(|x|\) becomes very large, \(\frac{3}{x+2}\) approaches 0. Therefore, \(f(x)\) approaches \(-1\), so the horizontal asymptote is \(y=-1\). The line \(x=-2\) is the vertical asymptote because the denominator becomes zero there. Exam tip: For a function of the form \(\frac{a}{x-h}+k\), the horizontal asymptote is directly \(y=k\).

The coefficient in (\frac{-2}{x-3}) is negative. So its branches are like (-\frac{1}{x}) in opposite quadrants.

The denominator \((x+2)^2\) becomes zero at \(x=-2\), so the function is undefined there. As \(x\) approaches \(-2\), the value of \(f(x)\) increases without bound; therefore, \(x=-2\) is the vertical asymptote. The option \(x=2\) results from a sign error, while \(y=0\) is the horizontal asymptote. Exam tip: For a reciprocal function, set the denominator equal to zero to find the vertical asymptote.

The range follows from the sign and limiting behavior of the expression. The domain excludes \(x=1\), and for every allowed \(x\), \((x-1)^2>0\). Hence \(2/(x-1)^2>0\), so \(f(x)=2/(x-1)^2-3>-3\). The value \(-3\) can never occur because the positive fraction can never equal zero. On the other hand, as \(x\) approaches 1, the fraction becomes arbitrarily large, so there is no upper bound. Every value greater than \(-3\) is achieved: for \(y>-3\), choose \((x-1)^2=2/(y+3)\). Thus the range is \((-3,\infty)\), making option A correct. Option B incorrectly includes the unattainable endpoint.

The direct answer is option A: \((-\infty,2)\). The function is \(f(x)=-\frac{3}{(x+1)^2}+2\). First, x cannot equal −1 because the denominator would be zero. For every allowed x, \((x+1)^2>0\). Dividing 3 by this positive number gives a positive value, and the negative sign makes \(-\frac{3}{(x+1)^2}<0\). Therefore f(x)=2−a for some a>0, so f(x)<2. The value 2 is never reached because the fraction can never be zero. As x becomes very large in either direction, the fraction approaches zero and f(x) approaches 2 from below, but never equals it. Hence the range is \((-\infty,2)\). Option A is correct. Option B, \((2,\infty)\), is wrong because every output is below 2. Option C, \((-\infty,2]\), incorrectly includes 2. Option D, \([2,\infty)\), has the wrong side and incorrectly includes 2. The domain restriction x≠−1 does not create a missing output other than the upper endpoint; all values less than 2 can occur. Remember: a negative positive fraction added to 2 gives values strictly less than 2.

Since \(x>4\), we have \(3x-12=3(x-4)>0\). The signum of a positive number is \(1\), so \(f(x)=1\). Note that the value would be \(0\) at \(x=4\), but that point is excluded by \(x>4\). Exam tip: first determine the sign of the expression inside the signum function.

When \(x>2\), we have \(2-x<0\). The signum function has value \(-1\) for a negative argument; therefore, \(f(x)=\operatorname{sgn}(2-x)=-1\). Exam tip: \(\operatorname{sgn}(y)\) equals \(-1\) for \(y<0\), \(0\) for \(y=0\), and \(1\) for \(y>0\).