Graphs of standard functions

For the constant function \(f(x)=3\), the value of \(y\) is \(3\) for every \(x\). Thus, all points have the form \((x,3)\), forming a horizontal line parallel to the x-axis. Option B is incorrect because a line parallel to the y-axis has a fixed \(x\)-coordinate. Exam tip: the graph of \(y=c\), where \(c\) is constant, is always parallel to the x-axis.

For \(y=|x|\), the smallest value of \(|x|\) occurs when \(x=0\). Then \(y=|0|=0\). So the graph touches at \((0,0)\), which is the vertex.

The basic parabola \(y=x^2\) has its vertex at \((0,0)\). Adding \(+2\) shifts the entire graph 2 units upward, so the new vertex is \((0,2)\). Option C incorrectly treats 2 as the x-coordinate. Exam tip: for \(y=(x-h)^2+k\), the vertex is \((h,k)\).

((x-3)) means shifting the graph (3) units to the right. In exams, read (x-a) as a right shift.

The governing concept is translation of the standard modulus graph y = |x|, whose vertex is (0, 0). In y = |x + 2|, the expression x is replaced by x + 2, so the graph shifts 2 units to the left without any vertical shift. Consequently, the vertex becomes (-2, 0). The same result follows algebraically: the sharp turning point occurs when the expression inside the modulus is zero, so x + 2 = 0 and x = -2; then y = |0| = 0. Thus option A is correct. The point (2, 0) would result from |x - 2|, while the other choices confuse a horizontal displacement with a y-coordinate.

For (\sqrt{x-4}), the smallest (x) is (4) and then (y=0). In exams, start a square-root graph by making the inside expression (0).

The governing concept is the asymptotic behavior of a rational function. A vertical asymptote occurs where the denominator becomes zero, provided the zero is not cancelled by a common factor. Here the denominator is x - 2, so solve x - 2 = 0 to obtain x = 2. As x approaches 2 from either side, the numerator remains 1 but the denominator approaches zero, causing the magnitude of the function to increase without bound. Thus x = 2 is the vertical asymptote and option B is correct. The equation y = 0 describes the horizontal asymptote because the function approaches zero as |x| becomes large. The other choices do not make the denominator vanish.

The horizontal asymptote of (\frac{1}{x}) is (y=0), and (+1) shifts it to (y=1). In exams, apply vertical shifts to asymptotes too.

In (y=\frac{1}{x}), (x) and (y) have the same sign. In exams, remember that the reciprocal graph lies in quadrants (1) and (3).

For (y=x^2), (y\ge 0) and the graph lies on both sides of the (y)-axis. In exams, identify the parabola by its (U)-shape.

The graph of (x^3) rises continuously from left to right. In exams, remember (x^3) is increasing on the whole real domain.

The graph of (|x|) decreases up to (x=0) and then increases. In exams, see the left arm of the (V)-graph as decreasing.

On the (x)-axis, set (y=0), giving (x^2-4=0). In exams, set the related coordinate to (0) for intercepts.

(x^2+1) is always at least (1), so (y=0) is impossible. In exams, use the minimum value to check (x)-intercepts.

The slope of (2x) is (2), greater than the slope (1) of (x). In exams, treat the coefficient of (x) as the slope in a linear function.

(-|x|) is always (0) or less and equals (0) at (x=0). In exams, the vertex of a downward (V)-graph is the maximum.

For (y=\sqrt{x}), (x\ge 0) and (y\ge 0). In exams, associate the square-root graph with the first quadrant.

The governing concept is that an intersection point must satisfy both function equations and all domain restrictions. Set x^2 = √x. Since √x is real only for x ≥ 0, restrict the search to nonnegative x. Squaring is then valid and gives x^4 = x, or x(x^3 - 1) = 0. Hence x = 0 or x = 1. Substituting into either function gives y = 0 and y = 1 respectively, so the intersections are (0, 0) and (1, 1). Therefore option C is correct. The point (-1, 1) is invalid because √(-1) is not real. Options A and B each omit one of the two valid solutions.

On the (y)-axis, (x=0), and only (x+5) gives (f(0)=5). In exams, set (x=0) for the (y)-intercept.

The minimum of (|x-1|) is (0), so the total minimum is (2). In exams, take the minimum of a modulus part as (0).

((x+1)^2) shifts the graph (1) unit left and (-3) downward. In exams, for (y=(x-h)^2+k), use vertex ((h,k)).

The governing concept is an even function. A graph is symmetric about the y-axis exactly when f(-x) = f(x) for every x in its domain. For f(x) = x^2, f(-x) = (-x)^2 = x^2 = f(x), so the parabola has mirror symmetry about the y-axis. Therefore option B is correct. In contrast, f(x) = x^3 is odd because f(-x) = -f(x), giving symmetry about the origin rather than the y-axis. The function x + 1 is neither even nor odd. The real function √x has domain x ≥ 0 only, so reflecting its graph across the y-axis would require points with negative x that are not in its domain.

(x^3) is an odd function, so its graph is symmetric about the origin. In exams, check (f(-x)=-f(x)).

(\frac{1}{x^2}) is always positive and (x\ne 0). In exams, when (x^2) is in the denominator, remember (y) is positive.

The direct answer is option D, 9. The graph y=x^2 means that the y-coordinate is found by squaring the given x-coordinate. For x=-3, y=(-3)^2=(-3)(-3)=9. Thus the graph contains the point (-3,9). Option D is correct. Option A, -9, forgets that the product of two negative numbers is positive. Option B, -6, is not the square of -3 and may come from multiplying by 2 or using an unrelated operation. Option C, 6, is also not obtained from the rule y=x^2. The important idea is that a square cannot be negative for a real number: x^2≥0. The minus sign is inside the square, so it disappears when the two equal negative factors are multiplied. Do not confuse (-3)^2 with -3^2; the former is 9, while the latter is conventionally -9 because the exponent applies before the leading minus. The supplied answer is consistent.