Graphs of standard functions

For \(y=\frac{1}{x}\), \(f(-x)=-f(x)\), so its graph is symmetric about the origin. The lines \(x=0\) and \(y=0\) are its asymptotes. Exam tip: \(1/x^2\) is symmetric about the y-axis, not the origin.

The direct answer is A, \((-2,-1)\). Start with the parent graph \(y=x^3\), which contains the point \((0,0)\). In \(y=f(x+2)-1\), replacing x by \(x+2
displays the graph 2 units left, while subtracting 1 outside the function moves it 1 unit down. Therefore \((0,0)\) becomes \((-2,-1)\). Option A matches both movements. Option B, \((2,-1)\), moves right instead of left. Option C, \((-2,1)\), moves up instead of down. Option D has unrelated coordinates and represents neither required translation. Memory cue: plus inside means left; minus outside means down.

For a graph y = f(x), replacing x by -x gives y = f(-x), which is the reflection of the original graph in the y-axis. Here f(x) = x^3, so y = f(-x) = (-x)^3. Reflection in the line y = x would produce the inverse relation y = ∛x, not the given graph. Exam tip: replacing x by -x causes reflection in the y-axis, whereas replacing y by -y causes reflection in the x-axis.

Since \(f(-x)=(-x)^5=-x^5=-f(x)\), the function \(f(x)=x^5\) is odd. The graph of an odd function is symmetric about the origin. Option B would describe an even function, but this graph is not symmetric about the y-axis. Exam tip: if \(f(-x)=-f(x)\), the graph has origin symmetry.

A function is even when \(f(-x)=f(x)\), and the graph of an even function is symmetric about the y-axis. Here, \(f(-x)=(-x)^6=x^6=f(x)\), so the graph has y-axis symmetry. Symmetry about the origin corresponds to an odd function, which this function is not. Exam tip: a power function involving an even power of \(x\) is symmetric about the y-axis.

The direct answer is option A, -3. The greatest integer function floor(x) means the greatest integer that is less than or equal to x. For -2.3, the integers near it are -3 and -2. Although -2 is numerically greater, it is not less than or equal to -2.3; in fact, -2>-2.3. The integer -3 satisfies -3≤-2.3, so floor(-2.3)=-3. Option A is correct. Option B, -2, is the number obtained by moving toward zero, but the greatest integer function is not ordinary truncation toward zero. Option C, 2, has the wrong sign and is not less than or equal to -2.3. Option D, 3, is also positive and cannot satisfy the required inequality. On a number line, the answer is the integer immediately to the left of -2.3. A common mistake is to think the answer is -2 because it is closest; closeness is not the rule. The supplied answer is correct.

The floor function \(\lfloor x\rfloor\) gives the greatest integer less than or equal to \(x\). For every number in the interval \(2\le x<3\), the greatest integer not exceeding that number is 2. This includes the left endpoint \(x=2\), where the value is exactly 2, but excludes \(x=3\), where the value would become 3. Therefore option A is correct.

For instance, \(\lfloor2\rfloor=2\), \(\lfloor2.4\rfloor=2\), and \(\lfloor2.99\rfloor=2\). Thus the function remains constant at 2 throughout the stated interval, so its graph is a horizontal segment at height 2. Option B would apply at \(x=3\), but that endpoint is not included.

The signum function \(\operatorname{sgn}(u)\) equals \(-1\) for \(u<0\) and \(1\) for \(u>0\), so its jump occurs where \(u=0\). Here, \(2x-8=0\), giving \(x=4\). At \(x=4\), the function value is \(0\), while the left- and right-hand limits are \(-1\) and \(1\), respectively. Exam tip: To locate the jump of a signum function, set its inner expression equal to zero.

The expression \(\min(x,4)\) selects the smaller of \(x\) and \(4\). When \(x>4\), the smaller value is always \(4\), so \(y=4\) and the graph is horizontal in this region. The option \(y=x\) is valid only for \(x\leq 4\). Exam tip: For a minimum function, compare the expressions separately on the relevant intervals.

When \(x\ge 3\), we have \(x-3\ge 0\), so \(|x-3|=x-3\). Therefore, \(y=(x-3)-(x-3)=0\). Option B would result from incorrectly using \(|x-3|=-(x-3)\). Exam tip: For an absolute-value function, first determine the sign of the expression inside the modulus.

\(y=x^3\) 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 even and is symmetric about the y-axis. Exam tip: test \(f(-x)=-f(x)\) for origin symmetry.

At an intersection, the two graphs have the same \(y\)-value, so \(|x+1|=5\). Thus, \(x+1=5\) or \(x+1=-5\), giving \(x=4\) or \(x=-6\). Therefore, option A is correct. Exam tip: for \(|u|=a\) with \(a>0\), always consider both cases \(u=\pm a\).

At an intersection, the two functions have the same \(y\)-value, so \(\sqrt{x}=x-2\). Since the left side is non-negative, we must have \(x-2\ge 0\), hence \(x\ge 2\). Squaring gives \(x=(x-2)^2\), or \(x^2-5x+4=0\), whose roots are \(x=1\) and \(x=4\). However, \(x=1\) violates the condition \(x\ge 2\) and is an extraneous root introduced by squaring. Therefore, \(x=4\); checking gives \(\sqrt{4}=2=4-2\). Exam tip: Always substitute the roots back into the original equation after squaring.

At an intersection, both functions have the same y-value, so 1/x = x, with x ≠ 0. Hence x² = 1, giving x = 1 or x = −1. Since y = x, the intersection points are (1, 1) and (−1, −1).

The direct answer is A: 3 units right and 2 units up. For a graph \(y=f(x)\), replacing x by \(x-h\) shifts it h units right, because the same function value is reached when x is increased by h. Thus \(f(x-3)\) moves the graph 3 units right. Adding 2 outside, as in \(f(x-3)+2\), raises every y-value by 2. Option A states both correct movements. Option B reverses the horizontal direction. Option C reverses the vertical direction. Option D swaps the numbers and is therefore incorrect. Exam cue: inside signs work oppositely; outside signs work normally.

A point (x, y) on y = f(x) becomes (x, −y) when the function values are negated, which reflects the graph across the x-axis. Adding 4 then changes each output to −y + 4, shifting the reflected graph 4 units upward.

With modulus outside, all (y)-values become non-negative. Hence the part below the (x)-axis reflects upward.

Because of (|x|), even for (x<0), the input to (f) is positive. Hence the right-side graph reflects to the left.

(\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}), but (x=2) remains excluded. So the graph is like (y=x+2) with a hole at (x=2).

For \(x>1\), both \(x-1\) and \(x+3\) are positive. Hence, \(|x-1|=x-1\) and \(|x+3|=x+3\). Therefore, \(y=(x-1)-(x+3)=-4\). Option B results from ignoring the negative sign, while the correct value is the constant \(-4\) for every \(x>1\). Exam tip: Always determine the sign of each expression before removing modulus bars.

The expression represents the sum of the distances of the point \(x\) from the fixed points \(-2\) and \(4\) on the number line. The distance between these points is \(|4-(-2)|=6\), so by the triangle inequality the minimum sum is 6. In fact, the sum remains 6 for every \(x\) in the interval \([-2,4]\). Exam tip: The minimum value of \(|x-a|+|x-b|\) is generally \(|a-b|\).

On (4\le x\le 10), the sum is ((x-4)+(10-x)=6). In a modulus distance sum the graph is horizontal between the two points.

Since \(f(-x)=|-x|=|x|\), \(|x|\) is an even function, so its graph is symmetric about the \(y\)-axis. Also, \(|x|\ge0\), giving range \([0,\infty)\). It is not one straight line because the slope changes at \(x=0\). Exam tip: test evenness using \(f(-x)=f(x)\).

On the \(x\)-axis, \(y=0\). Hence \(|2x+6|-4=0\), so \(|2x+6|=4\). This gives two cases: \(2x+6=4\), yielding \(x=-1\), and \(2x+6=-4\), yielding \(x=-5\). Therefore, the correct values are \(x=-5,-1\). Exam tip: for an equation \(|u|=a\), always consider both \(u=a\) and \(u=-a\). Option B does not result from correctly solving the two cases.

For a point on the \(x\)-axis, \(y=0\). Thus, \(|x^2-1|=0\), which is possible only when \(x^2-1=0\). Hence \(x=\pm1\), giving the points \((-1,0)\) and \((1,0)\). The graph touches the \(x\)-axis at these points rather than crossing it. Exam tip: To find intersections with the \(x\)-axis, set \(y=0\).