Graphs of standard functions

The governing concept is the piecewise definition of the absolute-value function. For every positive x, |x|=x. Substitution into the expression gives y=|x|−x=x−x=0. Thus all points corresponding to x>0 have ordinate zero, and this part of the graph lies along the x-axis. Option A is therefore correct. Option B would result from the different expression |x|+x, not the given subtraction. Option C would require an incorrect replacement |x|=−x for positive x; that replacement applies when x<0. Option D ignores the second term altogether. At x=0 the function also has value zero, although the question specifically asks about the positive-x portion.

The governing concept is a transformation of the input variable. Replacing x by −x changes every point (x,y) on y=f(x) to (−x,y) on y=f(−x). The y-coordinate remains unchanged, while the x-coordinate changes sign. This is precisely a reflection in the y-axis. Hence option A is correct. Replacing the output y by −y, or writing y=−f(x), would produce reflection in the x-axis instead. An upward shift would require an expression such as f(x)+k with k>0, and a right shift would normally use f(x−h), not f(−x). Thus the negative sign inside the function causes a horizontal, not vertical, reflection.

The governing concept is the vertical asymptote of a rational function. A vertical asymptote occurs at a value of x where the denominator becomes zero, provided the numerator does not cancel that factor. For y = 1/(x − 4), set the denominator equal to zero: x − 4 = 0, so x = 4. As x approaches 4 from either side, the denominator approaches zero and the function values become unbounded, confirming the vertical asymptote. The numerator is the nonzero constant 1, so no cancellation occurs. x = −4 would make the denominator −8, not zero; y = 4 is a horizontal line, and y = 0 is the horizontal asymptote, not the vertical one.

The governing concept is the range of a quadratic in vertex form. For every real x, (x-1)^2 is non-negative, so -(x-1)^2 is at most 0. Consequently y = 4-(x-1)^2 is always less than or equal to 4. Equality occurs when x-1 = 0, that is, at x = 1; the vertex is therefore (1,4) and gives the maximum value 4. As |x-1| grows, the squared term grows without bound and y decreases without bound. Hence the range is (-∞,4], making option A correct. Option B reverses the inequality, option C uses the wrong vertex value, and option D incorrectly includes values greater than 4.

A transformation applied outside a function changes its output values. Starting with y=x^3, replacing y by -y gives y=-x^3, so every point (x,y) is mapped to (x,-y). This reverses the sign of every vertical coordinate while leaving x unchanged, which is precisely reflection in the x-axis. Therefore option A is correct. Reflection in the y-axis would replace x by -x and produce y=(-x)^3=-x^3 as well for this particular odd function, but the direct transformation shown by the outside negative sign is an x-axis reflection; the question asks how the displayed expression is obtained. A shift would alter coordinates by a constant and cannot account for the sign reversal.

Factor the negative sign from the inside expression: -x-2 = -(x+2). Hence f(-x-2) can be written as f(-(x+2)), which is option A. This form makes the transformation structure visible: the negative applied to the input indicates reflection in the y-axis, while the addition of 2 inside the brackets corresponds to a horizontal shift when the transformation is interpreted carefully. Option B changes the sign of the shift and omits the reflection, option C omits the reflection entirely, and option D represents -(x-2)=-x+2, which is a different expression. Algebraic factoring must therefore be done before interpreting the graph.

The governing idea is the test for even symmetry: a function is symmetric about the y-axis when f(−x) = f(x) for every x in its domain. Here f(x) = √|x|, so f(−x) = √|−x| = √|x| = f(x), because the absolute value removes the sign of x. Also, |x| is defined for every real x, so the domain is all real numbers, not only x ≥ 0. Since a square root is never negative, the range is [0, ∞), not (−∞, 0]. Thus the graph has two reflected branches on either side of the y-axis, making option A correct. Origin symmetry would require f(−x) = −f(x), which does not hold except at zero.

Answer: option A, parallel. In slope-intercept form y=mx+c, m is the slope and c is the y-intercept. For y=2x+1, the slope is 2 and the intercept is 1. For y=2x−7, the slope is also 2, but the intercept is −7. Two distinct lines with equal slopes have the same direction and are parallel. They are not the same line because their y-intercepts are different. They are not perpendicular: for non-vertical perpendicular lines, the product of their slopes is −1, whereas here 2×2=4. They do not intersect because equal slopes and unequal intercepts make the equations inconsistent when equated. Thus A is the only correct option. Memory cue: same slope plus different intercept means parallel; same slope plus same intercept means the same line.

The governing concept is the standard equation of a circle centered at the origin: x^2 + y^2 = r^2. Starting with y = √(25 − x^2), square both sides to obtain y^2 = 25 − x^2. Rearranging gives x^2 + y^2 = 25, so the radius is √25 = 5 and the center is (0, 0). The original square-root form also imposes y ≥ 0, because a principal square root is never negative. Therefore it represents only the upper semicircle of that circle, not the complete circle. Option A is correct. Option B would describe radius √5, not 5; option C is a straight line; and option D is a parabola. The domain is −5 ≤ x ≤ 5, consistent with this semicircle.

The parent graph y = 1/x^2 is symmetric about the y-axis, whose equation is x = 0. Replacing x by x + 1 shifts the graph one unit to the left, so its vertical symmetry axis shifts from x = 0 to x = −1. This can also be checked algebraically: for any non-zero t, f(−1 + t) = 1/t^2 and f(−1 − t) = 1/(−t)^2 = 1/t^2. Thus points equally distant from x = −1 have equal function values, proving reflectional symmetry about that line. The line y = 0 is only a horizontal asymptote, while x = 1 and y = −1 do not preserve the graph. Hence option A is correct.

For every real x, x^2 ≥ 0, so x^2 + 9 ≥ 9. The denominator is positive and has its smallest possible value, 9, when x = 0. For positive numbers, the reciprocal decreases as the denominator increases; therefore 1/(x^2 + 9) ≤ 1/9. Equality occurs at x = 0, giving y = 1/9, so option A is the maximum value. Option B confuses the denominator with its reciprocal. Option C is approached as |x| becomes very large but is never attained for a finite real x because the numerator is 1. Option D is impossible because x^2 + 9 is never equal to 1. Thus the inequality and equality condition establish the answer unambiguously.

A multiplier applied inside the input causes a horizontal transformation. Suppose (u, f(u)) is a point on the original graph y = f(x). For the new graph y = f(3x) to have the same output f(u), we require 3x = u, so x = u/3. The new point is therefore (u/3, f(u)); its horizontal distance from the y-axis becomes one-third of the original distance. Consequently, the graph is compressed horizontally toward the y-axis by a factor of 3. The output values do not change, so there is no vertical stretch. A right shift would involve f(x − 3), and an upward shift would involve f(x) + 3. Thus option A is correct.