Graphs of standard functions

The basic function \(y=|x|\) has vertex \((0,0)\). In \(|x-3|\), the graph shifts 3 units to the right, and the outside \(+2\) shifts it 2 units upward. Therefore, the vertex is \((3,2)\). Option B reverses the horizontal shift, while option C reverses the vertical shift. Exam tip: for \(y=|x-h|+k\), the vertex is directly \((h,k)\).

Since \((x+2)^2\ge 0\), we have \(-(x+2)^2\le 0\), so \(f(x)=-(x+2)^2+5\le 5\). Equality occurs when \(x=-2\), hence the maximum value is 5. Note that \(-2\) is the x-coordinate of the vertex, not the maximum value of the function. Exam tip: for a downward-opening parabola, the y-coordinate of the vertex gives the maximum value.

For real values, (7-x\ge 0), so (x\le 7). The expression inside a square root must not be negative.

The denominator is zero when \(x+3=0\), so \(x=-3\). Since the numerator is nonzero there, the function has a vertical asymptote at \(x=-3\).

As \(|x|\) becomes very large, \(\frac{2}{x-1}\) approaches 0. Therefore, \(f(x)\) approaches 4, so the horizontal asymptote is \(y=4\). The line \(x=1\) is the vertical asymptote, not the horizontal one. Exam tip: For a function in the form \(\frac{a}{x-h}+k\), the horizontal asymptote is \(y=k\).

The modulus \(|2x-6|\) is minimized when its inner expression is zero, because the least possible value of an absolute value is 0. Thus, \(2x-6=0\), giving \(x=3\). For example, at \(x=6\), the function value is 6, so it cannot be the minimum. Exam tip: For a function of the form \(|ax+b|\), set \(ax+b=0\) to find the minimum point.

At an x-intercept, the function value is zero. Thus, \(|x+1|-4=0\), so \(|x+1|=4\). Therefore, \(x+1=4\) or \(x+1=-4\), giving \(x=3\) or \(x=-5\). Hence, the x-intercepts are \((3,0)\) and \((-5,0)\). Exam tip: For a modulus function, set \(f(x)=0\) and solve both positive and negative cases.

At any point on the x-axis, \(y=f(x)=0\). Thus, \((x-3)^2-4=0\), so \((x-3)^2=4\). Hence \(x-3=\pm2\), giving \(x=1\) or \(x=5\). Therefore, the x-intercepts are \((1,0)\) and \((5,0)\). Exam tip: To find x-intercepts, set the function equal to zero.

In \(f(x)=3|x-2|\), the outside coefficient 3 multiplies every \(y\)-value by 3. Hence the graph is vertically stretched by a factor of 3, while its vertex remains at \((2,0)\). Option B describes vertical compression, which occurs when the outside coefficient lies between 0 and 1. Exam tip: a coefficient outside the modulus causes vertical scaling, whereas a change inside the modulus causes horizontal shifting.

In \(f(x)=\frac{1}{2}x^2\), every \(y\)-value of the parent function \(y=x^2\) is multiplied by \(\frac{1}{2}\). Thus, the graph is vertically compressed toward the \(x\)-axis and the parabola appears wider. It is not shifted left or down because there is no change inside \(x\) and no constant term has been added. Exam tip: when \(0<a<1\), the graph of \(y=ax^2\) is vertically compressed.

The negative sign reflects in the (x)-axis, and (2) gives a vertical stretch. So the parabola opens downward and becomes narrower.

(\sqrt{x+2}) starts when (x+2=0), that is (x=-2). The outside (-3) makes the starting point ((-2,-3)).

Since (\sqrt{5-x}\ge 0), (f(x)\ge 2). The square root can grow as (x) decreases, so the range is ([2,\infty)).

(\frac{1}{(x+1)^2}>0), so the function is greater than (3) but never equal to (3). Hence the range is ((3,\infty)).

(\frac{1}{(x-3)^2}) is positive, so its negative is always less than (0). It never becomes (0).

When \(x=5\), we get \(x-5=0\). By definition of the signum function, \(\operatorname{sgn}(0)=0\); hence, \(f(5)=0\). The value 1 applies when \(x-5>0\), while -1 applies when \(x-5<0\). Exam tip: Always check the zero, positive, and negative cases separately for a signum function.

When \(x<4\), we have \(4-x>0\). The signum function has value \(1\) for a positive argument, so \(f(x)=\operatorname{sgn}(4-x)=1\). The value 0 occurs only at \(x=4\), while -1 occurs when \(x>4\). Exam tip: remember that \(\operatorname{sgn}(y)=1\) for \(y>0\), 0 for \(y=0\), and -1 for \(y<0\).

The governing concept is the greatest-integer, or floor, function. For any real number \(t\), \(\lfloor t\rfloor\) is the greatest integer less than or equal to \(t\). First substitute the given input: \(x+2=1.6+2=3.6\). The greatest integer not exceeding 3.6 is 3, so \(f(1.6)=\lfloor3.6\rfloor=3\). Therefore option A is correct. The answer is not 3.6 because the floor function removes the fractional part by moving downward to the nearest integer. Option B would be obtained by ignoring the added 2, option C by rounding 3.6 upward, and option D by failing to apply the function.

The direct answer is A, 2. First substitute x=1.4 into the inside expression: \(2x=2(1.4)=2.8\). The floor symbol \(\lfloor t\rfloor\) means the greatest integer that is less than or equal to t. Since 2 is the greatest integer not exceeding 2.8, \(\lfloor2.8\rfloor=2\). Option A is correct. Option B, 3, is wrong because 3 is greater than 2.8, so it is not allowed by the floor definition. Option C, 2.8, is the value before applying the floor operation, not the function value. Option D, 1, is too small and is not the greatest integer at most 2.8. For a positive decimal, floor means remove the decimal part, but always check the definition, especially for negative numbers.

For every \(x\) in \([3,4)\), we have \(3\le x<4\), so the greatest integer less than or equal to \(x\) is \(\lfloor x\rfloor=3\). Therefore, \(f(x)=3\), giving a horizontal graph segment at \(y=3\) throughout this interval. Option B is incorrect because \(x=4\) is excluded from the interval; the value becomes 4 only at the next step. Exam tip: for the greatest integer function, \(f(x)=n\) on every interval \([n,n+1)\).

The graph of \(y=x^2\) is a parabola symmetric about the y-axis, with its minimum at \((0,0)\) and a horizontal tangent there. Although \(y=|x|\) is also symmetric, it has a sharp corner at the origin. Exam tip: a smooth U-shape indicates a parabola.

At any point on the x-axis, the y-coordinate is 0. Hence, \(x^2-4x=0\), which factors as \(x(x-4)=0\). Therefore, \(x=0\) or \(x=4\), so option A is correct. Option B incorrectly includes \(-4\), which is not a zero of this function. Exam tip: To find the x-intercepts of a quadratic function, set \(f(x)=0\).

At every point on the x-axis, the y-coordinate is 0. Hence, set \(f(x)=0\): \(x^3-8=0\), so \(x^3=8\) and \(x=2\). Therefore, the x-coordinate of the intercept is 2. Choosing \(x=8\) incorrectly treats the constant term as the answer. In an exam, set \(y=0\) to find an x-intercept.

The parent function \(y=x^3\) has its point of inflection at \((0,0)\). In \((x+3)^3-1\), the graph is shifted 3 units left and 1 unit down, so the new point of inflection is \((-3,-1)\). Alternatively, \(f''(x)=6(x+3)\) gives \(x=-3\), and \(f(-3)=-1\). Option B incorrectly treats \(x+3\) as a shift to the right. Exam tip: the point of inflection of \((x-h)^3+k\) is \((h,k)\).

The direct answer is option A: x=y^2 does not represent y as a function of x. For example, x=4 gives y^2=4, so y=2 or y=-2. One input x therefore has two outputs, violating the definition of a function. Equivalently, a vertical line such as x=4 cuts the graph twice, so the vertical-line test fails. Option A is correct. Option B, y=x^2, gives exactly one y for every x. Option C, y=|x|, also gives one output for each x. Option D, y=\sqrt{x}, is defined for x\ge0 and has one non-negative output. Common mistake: confusing the vertical-line test with the horizontal-line test.