Graphs of standard functions

For real-valued outputs, the expression inside the square root must be non-negative: \(x-4\ge 0\), so \(x\ge 4\). At \(x=4\), \(f(4)=0\), giving the starting point \((4,0)\). Therefore, the graph begins at \(x=4\), not at any smaller value. Exam tip: the graph of \(\sqrt{x-a}\) starts at \(x=a\).

The direct answer is option B: the graph lies in the first and third quadrants. Since y=1/x, x and y have the same sign: if x>0, then y>0, placing the point in quadrant I; if x<0, then y<0, placing it in quadrant III. x=0 is not allowed, and y=0 is never obtained, so the graph does not lie on the axes. Option B is correct. Option A misses the entire negative branch. Option C would be correct for y=-1/x, where the signs are opposite. Option D is incomplete because the graph also has a first-quadrant branch. Exam cue: for xy=1, equal signs give quadrants I and III.

The governing concept is symmetry of an even function. For f(x)=x^2, compare opposite inputs: f(-x)=(-x)^2=x^2=f(x). This equality means that points with x-coordinates x and -x have the same y-coordinate, so the graph is mirror-symmetric about the vertical line x=0. The line x=0 is precisely the y-axis, making option B correct. The x-axis is not an axis of symmetry because reflecting a point across it changes y to -y. The line y=x is not generally a symmetry line for this parabola, and x=1 would be the axis for a translated parabola such as (x-1)^2, not for x^2. Thus the even-function test confirms the answer.

Since (f(-x)=(-x)^3=-x^3=-f(x)), it is symmetric about the origin. Identify odd functions by (f(-x)=-f(x)).

For the given function, \(y=|x|\). Substituting \(x=-5\) gives \(y=|-5|=5\). An absolute value represents distance from zero, so it is never negative. The option \(-5\) incorrectly gives the value of \(x\) without applying the modulus. Exam tip: \(|a|\) is always the non-negative value of \(a\).

Substituting \(x=9\) into the function gives \(y=\sqrt{9}=3\). The principal value of a square root is always non-negative, so \(-3\) is not correct. Exam tip: remember that \(\sqrt{a^2}=|a|\), not always \(a\).

Substituting \(x=-2\) into the function gives \(y=(-2)^2=4\). Hence, the correct value is 4, and the corresponding point is \((-2,4)\). The value is not \(-4\) because the entire number \(-2\) is being squared; use brackets when squaring a negative number.

The governing concept is finding a y-intercept. Every point on the y-axis has x-coordinate 0, so substitute x=0 into the function. We obtain f(0)=0^2-4=0-4=-4. Therefore the graph meets the y-axis at the point (0,-4), and the requested y-coordinate is -4, making option B correct. The value 4 results from losing the negative sign. The value 0 is the squared part before subtracting 4, not the final function value. The value 2 is unrelated to the y-intercept; it may arise from confusing this question with solving a different equation. Remember that y-intercepts are found by setting x equal to zero, whereas x-intercepts require setting y, or f(x), equal to zero.

On the x-axis, \(y=0\). Therefore, \(x^2-4=0\), so \(x^2=4\) and \(x=2\) or \(x=-2\). Hence, the intercepts are \((2,0)\) and \((-2,0)\). In option C, substituting \(x=\pm1\) gives \(f(x)=-3\), so those values do not lie on the x-axis. Exam tip: To find x-intercepts, set \(f(x)=0\).

The direct answer is A: the graph cuts the x-axis 0 times. A point on the x-axis has y=0. For this graph, y=x^2+4. Since \\(x^2\ge0\\) for every real x, we have \\(x^2+4\ge4\\). Thus y can never be 0, so there is no x-intercept. A is correct. Option B would require one real solution of \\(x^2+4=0\\), but this gives \\(x^2=-4\\), impossible for real x. Option C would mean two x-intercepts; although a usual upward quadratic may have two, this one is shifted four units above the x-axis and has none. Option D has no algebraic or graphical basis. The vertex is \\( (0,4)\\), and the entire parabola lies above the x-axis. Memory cue: to count x-axis crossings, set y=0 and count real solutions.

Putting (x=0) gives (y=0), so the line (y=-x) passes through the origin. Check simple points first in a linear function.

The direct answer is B: all real x. The function f(x)=2 is constant: whatever real input x is chosen, the output remains 2. There is no square root, denominator, logarithm, or other restriction in the formula. Therefore every real number can be used as an input, and the domain is ℝ. Option A, only x=2, confuses the output value 2 with the allowed input values; x=2 is allowed, but it is not the only input. Option B is correct because values such as x=-3, 0, 1/2, and 10 all give f(x)=2. Option C, only x≥0, wrongly excludes negative real inputs even though the rule works for them. Option D, only x≠0, wrongly excludes x=0; f(0)=2 is perfectly defined. The graph is the horizontal line y=2, extending in both directions, which also shows that its domain is all real numbers. Memory cue: in a constant function, the fixed number is the output, not a restriction on x.

In \(f(x)=x+1\), 1 is added to every \(y\)-value of the original function \(y=x\), while the \(x\)-values remain unchanged. Therefore, the entire graph shifts 1 unit upward. Exam tip: \(f(x)+k\) shifts a graph upward, whereas \(f(x-k)\) produces a horizontal shift to the right.

The minimum of (|x|) is (0), so the minimum of (|x|+3) is (3). Keep the vertical shift in mind.

For every real \(x\), \(x^2\ge 0\). Hence \(f(x)=x^2+2\ge 2\), and at \(x=0\), \(f(0)=2\). Therefore, the minimum value is 2. Remember that 0 is the minimum of \(x^2\), but adding 2 shifts the minimum value of the function to 2.

On the line \(y=0\), the y-coordinate of every point is zero, so all its points have the form \((x,0)\) and lie on the x-axis. Therefore, the correct answer is the x-axis. Remember that the equation of the y-axis is \(x=0\), whereas \(x=1\) is a vertical line parallel to the y-axis.

Since \(g(x)=f(x)+5\), for every \(x\), the value of \(g(x)\) is 5 greater than the corresponding value of \(f(x)\). Thus, the parabola keeps the same shape and horizontal positions but shifts 5 units upward. Option B would represent \(f(x)-5\), which shifts the graph downward. Exam tip: \(y=f(x)+k\) shifts a graph upward by \(k\) units, while \(y=f(x)-k\) shifts it downward.

A point lies on the graph only if it satisfies \(y=|x|\). For \((-4,4)\), \(|-4|=4\), so it is correct. \((-4,-4)\) and \((4,-4)\) are incorrect because \(|x|\) can never be negative, while \((0,4)\) is incorrect since \(|0|=0\), not 4. Exam tip: substitute the point’s \(x\)-coordinate into the function and compare the result with its \(y\)-coordinate.

(\lfloor 2.7\rfloor=2), because it is the greatest integer less than or equal to (2.7). Remember the step-like shape of this graph.

In (\frac{1}{x^2}), the denominator is (x^2), so at (x=0) the denominator becomes zero. In fractional functions, check the denominator first.

The graph of (-|x|) is the reflection of (|x|) in the (x)-axis, and the vertex remains ((0,0)). The negative sign makes the (V)-shape open downward.

The governing concept is the piecewise definition of the signum, or sign, function. For a real number x, sgn(x)=1 when x>0, sgn(x)=0 when x=0, and sgn(x)=-1 when x<0. Since the question explicitly states x>0, the applicable case is the first one, so f(x)=1 and option A is correct. The value 0 belongs only to the input x=0, while -1 belongs to negative inputs. The expression x is not the output rule here; signum records only the sign of the input, not its magnitude. On its graph, all positive x-values correspond to the horizontal line y=1, which gives the same conclusion visually.

The governing concept is the signum function, which records whether its input is positive, zero, or negative. Its standard definition is sgn(x)=1 when x>0, sgn(x)=0 when x=0, and sgn(x)=−1 when x<0. Because the question gives x<0, the input is negative and the output must be −1. Therefore option C is correct. Option A belongs to the positive case x>0, while option B belongs only to the zero case x=0. Option D, x², is not a value assigned by the signum definition; it describes a different function and would also vary with x. The answer depends only on the sign of x, not on its magnitude, so every negative input produces the same value −1.

The governing concept is the piecewise definition of the signum, or sign, function. It is given by sgn(x) = 1 when x > 0, sgn(x) = 0 when x = 0, and sgn(x) = -1 when x < 0. Since the question specifically gives x = 0, the middle case applies directly: f(0) = sgn(0) = 0. Therefore option B is correct. The value 1 belongs to positive inputs, while -1 belongs to negative inputs. The function is not undefined at zero under its standard definition; zero is deliberately assigned as its output. Graphically, the two horizontal branches lie at y = 1 and y = -1, with the origin included at y = 0.

(\lfloor 3.2\rfloor=3), because it is the greatest integer less than or equal to (3.2). The greatest integer graph is step-like.