Graphs of Functions: Problems with Solutions

Solution:
For [tex]x\lt 0[/tex] the value of the function is positive:
[tex]y=-\frac{2}{x}=\frac{2}{-x}=\frac{2}{\left|x\right|}\gt 0[/tex]
So every point of the graph has a negative abscissa and a positive ordinate, i.e. it lies in quadrant [tex]\text{II}[/tex]. A table of values:
[tex]\begin{array}{c|c|c|c|c} x & -4 & -2 & -1 & -\frac{1}{2} \\ \hline y & \frac{1}{2} & 1 & 2 & 4 \end{array}[/tex]
The graph is one branch of the hyperbola [tex]y=-\frac{2}{x}[/tex]: as [tex]x[/tex] approaches [tex]0[/tex] from the left, [tex]y[/tex] grows without bound, and as [tex]x\to -\infty[/tex], [tex]y[/tex] approaches [tex]0[/tex]; the axes are never reached.
The whole hyperbola lies in quadrants [tex]\text{II}[/tex] and [tex]\text{IV}[/tex]; the condition [tex]x\lt 0[/tex] keeps only the branch in quadrant [tex]\text{II}[/tex]. The minus sign in front of the fraction does not make [tex]y[/tex] negative here, because [tex]x[/tex] itself is negative.

Solution:
The axis [tex]Oy[/tex] is the line [tex]x=0[/tex]. If the graph of [tex]y=f(x)[/tex] had two different points [tex]\left(0, y_1\right)[/tex] and [tex]\left(0, y_2\right)[/tex] on it, with [tex]y_1\ne y_2[/tex], the function would take two different values at the same argument [tex]x=0[/tex]. This contradicts the definition of a function: every [tex]x[/tex] from the domain has exactly one value [tex]f(x)[/tex]. In general, every vertical line [tex]x=c[/tex] meets the graph of a function in at most one point.
The other three lines can have two common points with a graph. For example, the graph of [tex]f(x)=2\left|x\right|-1[/tex] meets them at the points
[tex]Ox:\ \left(-\frac{1}{2}, 0\right),\ \left(\frac{1}{2}, 0\right)[/tex]
[tex]y=x:\ \left(-\frac{1}{3}, -\frac{1}{3}\right),\ (1, 1)[/tex]
[tex]y=-x:\ (-1, 1),\ \left(\frac{1}{3}, -\frac{1}{3}\right)[/tex]
while it meets [tex]Oy[/tex] only at [tex](0, -1)[/tex]. The answer is the axis [tex]Oy[/tex].

Solution:
The slopes of [tex]f_1[/tex] and [tex]f_2[/tex] are different ([tex]-2\ne 1[/tex]), so their graphs meet in exactly one point. We find it:
[tex]1-2x=x+4\iff -3x=3\iff x=-1,\quad y=f_1(-1)=1+2=3[/tex]
So the common point of the three graphs can only be [tex]P(-1, 3)[/tex], hence [tex]f_3(-1)=3[/tex]:
[tex]-(-1)+b=3\iff 1+b=3\iff b=2[/tex]
Check: [tex]f_3(x)=-x+2[/tex] gives [tex]f_3(-1)=3[/tex], so all three lines pass through [tex]P(-1, 3)[/tex].

Solution:
Since [tex]A(0, -2)[/tex] lies on the graph of [tex]f[/tex], we get [tex]f(0)=b=-2[/tex]. Parallel lines have equal slopes, so [tex]a=2[/tex] (and [tex]-2\ne -4[/tex], so the lines do not coincide). Hence
[tex]f(x)=2x-2[/tex]
The points where the two lines meet the axes are:
[tex]f(x)=2x-2:\quad (1, 0),\ (0, -2)[/tex]
[tex]g(x)=2x-4:\quad (2, 0),\ (0, -4)[/tex]
The figure is the trapezoid (the lines are parallel) with vertices [tex](1, 0)[/tex], [tex](2, 0)[/tex], [tex](0, -4)[/tex], [tex](0, -2)[/tex] in quadrant [tex]\text{IV}[/tex]. Its area is the area of the right triangle cut off from the axes by [tex]g[/tex] minus the area of the right triangle cut off by [tex]f[/tex]:
[tex]S=\frac{1}{2}\times 2\times 4-\frac{1}{2}\times 1\times 2=4-1=3[/tex]

Solution:
Since [tex]A[/tex] lies on [tex]y=ax[/tex] and its abscissa is [tex]1[/tex], its ordinate is [tex]a\times 1=a[/tex], so [tex]A(1, a)[/tex]. The foot of the perpendicular is [tex]A_1(1, 0)[/tex].
The triangle [tex]OA_1A[/tex] is right-angled at [tex]A_1[/tex], with legs [tex]OA_1=1[/tex] and [tex]A_1A=\left|a\right|[/tex] (a length is never negative: for [tex]a\lt 0[/tex] the point [tex]A[/tex] lies below [tex]Ox[/tex] and the leg is [tex]-a[/tex]). Therefore
[tex]S=\frac{1}{2}\times OA_1\times A_1A=\frac{1}{2}\times 1\times \left|a\right|=\frac{\left|a\right|}{2}[/tex]
The answer [tex]\frac{a}{2}[/tex] is correct only for [tex]a\gt 0[/tex]; for [tex]a\lt 0[/tex] it would give a negative area.

Solution:
Moving the graph [tex]2[/tex] units to the left means that [tex]g[/tex] takes at [tex]x[/tex] the value that [tex]f[/tex] takes at [tex]x+2[/tex]. We substitute [tex]x+2[/tex] for [tex]x[/tex] in the formula of [tex]f[/tex]:
[tex]g(x)=f(x+2)=\frac{1}{3}(x+2)-5=\frac{1}{3}x+\frac{2}{3}-5=\frac{1}{3}x-\frac{13}{3}[/tex]
Check: the points [tex](0, -5)[/tex] and [tex](3, -4)[/tex] of the graph of [tex]f[/tex] move to [tex](-2, -5)[/tex] and [tex](1, -4)[/tex], and indeed
[tex]g(-2)=-\frac{2}{3}-\frac{13}{3}=-5,\quad g(1)=\frac{1}{3}-\frac{13}{3}=-4[/tex]
Typical mistakes: [tex]\frac{1}{3}x-3=f(x)+2[/tex] is a shift up by [tex]2[/tex], and [tex]\frac{1}{3}x-\frac{17}{3}=f(x-2)[/tex] is a shift to the right.

Solution:
The graph of [tex]y=f(x-5)[/tex] is obtained from the graph of [tex]y=f(x)[/tex] by a translation of [tex]5[/tex] units to the right along [tex]Ox[/tex]: the point [tex]\left(t, f(t)\right)[/tex] goes to [tex]\left(t+5, f(t)\right)[/tex]. A horizontal translation changes only the abscissas of the points, not their ordinates, so the set of values does not change.
Formally: when [tex]x[/tex] runs through the domain of [tex]f(x-5)[/tex], the number [tex]t=x-5[/tex] runs through the whole domain of [tex]f[/tex], so [tex]f(x-5)[/tex] and [tex]f(t)[/tex] take exactly the same values. Hence the range of [tex]f(x)[/tex] is also [tex]\left[-2, 3\right][/tex].
Answers such as [tex]\left[3, 8\right][/tex] or [tex]\left[-7, -2\right][/tex] confuse the horizontal shift with a vertical one.

Solution:
In one coordinate system draw the graph of [tex]y=\left|x\right|[/tex] and the line [tex]y=-x[/tex] (the bisector of the second and fourth quadrants). The graph of [tex]y=\left|x\right|[/tex] consists of two rays with common end point [tex]O(0, 0)[/tex]:
[tex]y=\left|x\right|=\left\{\begin{array}{ll}x, & x\ge 0\\ -x, & x\lt 0\end{array}\right.[/tex]
For [tex]x\le 0[/tex] the left ray lies on the line [tex]y=-x[/tex], so the two graphs coincide along this whole ray, including the point [tex]O[/tex]. For [tex]x\gt 0[/tex] we have [tex]\left|x\right|=x\gt 0\gt -x[/tex]: the graph of [tex]y=\left|x\right|[/tex] is above the [tex]x[/tex]-axis and the line is below it, so there are no common points.
Hence the solutions are all [tex]x\le 0[/tex] (including [tex]x=0[/tex]). This agrees with the definition of absolute value: [tex]\left|x\right|=-x[/tex] exactly when [tex]x\le 0[/tex].
[tex]x\in\left(-\infty , 0\right][/tex]

Solution:
Table of values:
[tex]\begin{array}{c|ccccc} x & 0 & 1 & 4 & 9 & 16\\ \hline y & -2 & -1 & 0 & 1 & 2\end{array}[/tex]
The graph is the graph of [tex]y=\sqrt{x}[/tex] shifted [tex]2[/tex] units down and restricted to [tex]0\le x\le 16[/tex]: an arc from [tex](0, -2)[/tex] to [tex](16, 2)[/tex] that crosses the [tex]x[/tex]-axis at [tex](4, 0)[/tex]. Since [tex]\sqrt{x}[/tex] is increasing, so is [tex]y=\sqrt{x}-2[/tex], and its least and greatest values are at the ends of the interval:
[tex]y_{\min}=y(0)=0-2=-2,\qquad y_{\max}=y(16)=\sqrt{16}-2=2[/tex]
The function is continuous, so it takes every value between [tex]-2[/tex] and [tex]2[/tex]. The range is [tex]\left[-2, 2\right][/tex]. (The interval [tex]\left[0, 4\right][/tex] is the range of [tex]\sqrt{x}[/tex] before the shift, and [tex]\left[-2, 14\right][/tex] comes from subtracting [tex]2[/tex] from [tex]16[/tex] instead of from [tex]\sqrt{16}[/tex].)

Solution:
The graph consists of three parts:
for [tex]x\lt -1[/tex], the part of the line [tex]y=4-x[/tex] going up to the left from the point [tex](-1, 5)[/tex]; this end point does not belong to the graph (hollow point), because [tex]x=-1[/tex] does not satisfy [tex]x\lt -1[/tex];
for [tex]-1\le x\lt 2[/tex], the horizontal segment [tex]y=3[/tex] from the filled point [tex](-1, 3)[/tex] to the point [tex](2, 3)[/tex]; in particular [tex]f(-1)=3[/tex], not [tex]5[/tex];
for [tex]x\ge 2[/tex], the ray of the line [tex]y=x+1[/tex] starting at the filled point [tex](2, 3)[/tex] and going up to the right.
Now intersect the graph with the horizontal line [tex]y=5[/tex]. On the first part:
[tex]4-x=5\Rightarrow x=-1[/tex]
This value does not satisfy [tex]x\lt -1[/tex] and is rejected: the line [tex]y=5[/tex] passes exactly through the hollow point [tex](-1, 5)[/tex]. On the second part [tex]f(x)=3\ne 5[/tex], so there are no solutions. On the third part:
[tex]x+1=5\Rightarrow x=4\ge 2[/tex]
so [tex]x=4[/tex] is a solution. The equation has exactly one solution.