Graphs of standard functions

For the function \(f(x)=x^2\), we have \(y=x^2\), whose graph is an upward-opening U-shaped parabola. Its vertex is at \((0,0)\), and it is symmetric about the y-axis. Therefore, it is not a step-like or broken line. Exam tip: In \(y=ax^2\), the parabola opens upward when \(a>0\).

The direct answer is option B: the line y=x passes through the first and third quadrants. If x>0, then y=x>0, so the point is in quadrant I. If x<0, then y<0, so the point is in quadrant III. The origin (0,0) is also on the line, but it belongs to no quadrant. Option B is correct. Option A incorrectly pairs positive x with negative y in quadrant II. Option C describes a line such as y=-x, where signs are opposite. Option D includes quadrant IV, but points there have x>0 and y<0, which cannot satisfy y=x. Memory cue: y=x has equal signs and slope 1.

A point \((x,y)\) lies on the graph only when it satisfies \(y=3x\). For \((1,3)\), \(3=3\times1\), so it is correct. In option B, the coordinates are reversed; for \(x=3\), \(y\) should be 9. In option C, \(x=2\) requires \(y=6\), and in option D, \(x=-1\) requires \(y=-3\). Exam tip: substitute the point’s x-coordinate into the function and compare the resulting y-value.

At every point on the x-axis, the y-coordinate is 0. Therefore, set f(x) = 0: x + 4 = 0, which gives x = -4. Hence, the x-coordinate of the x-intercept is -4. Option A results from a sign error. Exam tip: To find the x-intercept, always set y or f(x) equal to 0.

At an \(x\)-intercept, the \(y\)-coordinate is zero. Thus, set \(f(x)=0\): \(2x-6=0\), which gives \(2x=6\) and \(x=3\). Therefore, the intercept point is \((3,0)\), and its \(x\)-value is 3. Exam tip: For an \(x\)-intercept, always set \(y=0\).

At an x-intercept, the y-coordinate is 0. Thus, set \(f(x)=0\): \(x^2-9=0\), which factors as \((x-3)(x+3)=0\). Hence \(x=3\) or \(x=-3\), so option B is correct. The complete intercept points are \((3,0)\) and \((-3,0)\). Exam tip: for x-intercepts, always substitute \(y=0\) or set \(f(x)=0\).

A graph cuts the x-axis at points where its y-coordinate is zero, so solve x² + 6 = 0. This gives x² = -6. For every real number x, x² is nonnegative, meaning x² ≥ 0, and therefore x² + 6 ≥ 6 > 0. There is no real solution and no point on the graph with y = 0. Consequently, the upward-opening parabola has vertex (0, 6) and remains entirely above the x-axis; it cuts the x-axis zero times. Option A is correct. Two intersections would occur for a quadratic with two real roots, while one would occur for a tangent double root, neither of which happens here.

At an \(x\)-intercept, the \(y\)-coordinate is \(0\). Thus, set \(f(x)=0\): \(|x|-5=0\), which gives \(|x|=5\). Therefore, \(x=5\) or \(x=-5\), so the intercepts are \((5,0)\) and \((-5,0)\). Exam tip: to find \(x\)-intercepts, always set the function value equal to zero.

The basic graph \(y=|x|\) has vertex \((0,0)\). The term \(|x+2|=|x-(-2)|\) shifts it 2 units left, and \(-3\) shifts it 3 units down. Therefore, the vertex is \((-2,-3)\). Exam tip: for \(y=|x-h|+k\), the vertex is \((h,k)\).

The graph of (\sqrt{x}) starts at ((0,0)) and shifts (1) unit up. Therefore the starting point is ((0,1)).

The direct answer is A: y=-|x|. Reflection in the x-axis sends every point (x,y) to (x,-y). Thus the x-coordinate stays the same, while the sign of the y-coordinate changes. The original graph is y=|x|, a V-shaped graph above or on the x-axis. After reflection, it becomes y=-|x|, a downward V whose values are non-positive. Option A is correct because replacing y by -y changes |x| into -|x|. Option B, y=|x|+1, moves the graph upward by 1; it is a vertical translation, not a reflection. Option C, y=|x-1|, moves the graph one unit right; it is a horizontal translation. Option D, y=x^2, is a different parabola and has no direct reflection relationship here. For a graph y=f(x), reflection in the x-axis is y=-f(x); reflection in the y-axis would instead be y=f(-x). Memory cue: x-axis reflection changes the output sign, while y-axis reflection changes the input sign.

For (\sqrt{x-3}), (x-3=0) gives (x=3), and the outside (2) is added. So the starting point is ((3,2)).

A horizontal asymptote describes the y-value that a function approaches as x tends to positive or negative infinity. For f(x) = 1/(x - 4), the denominator becomes arbitrarily large in magnitude when |x| becomes large, while the numerator remains 1. Hence 1/(x - 4) approaches 0, although it never equals 0 for any finite x. Therefore the horizontal asymptote is y = 0, so option A is correct. The equation x = 4 is instead the vertical asymptote because the denominator vanishes there. The choices y = 4 and x = 0 do not follow from the reciprocal form and confuse the shift or the type of asymptote.

The graph is the line \(y=x\), which has the form \(y=mx+c\) with \(m=1\). Equivalently, it rises by 1 unit for each 1-unit increase in \(x\), so its slope is 1. Therefore, option A is correct.

The graph of f(x)=5 is a horizontal line because the output remains 5 for every input x. The y-intercept is the point where the graph crosses the y-axis, and every point on that axis has x=0. Substituting x=0 into the function gives y=f(0)=5. Therefore the graph crosses the y-axis at the point (0,5), and the requested intercept value is 5.

Option B is correct. The number 0 identifies the x-coordinate of the intercept, but the y-intercept itself is usually reported by its y-value, 5, or by the point (0,5). The constant value of a constant function is its y-intercept whenever the function is defined at x=0, as it is here. The supplied answer and explanation are accurate.

The governing concept is the coordinate form of horizontal and vertical lines. In y = -2, the y-coordinate remains fixed at -2, while x may take every real value. Thus the points on the line are (x, -2), and the line extends horizontally from left to right. Every horizontal line is parallel to the x-axis, so option A is correct. Its slope is 0, which also confirms the conclusion. A line parallel to the y-axis must be vertical and have a constant x-coordinate, such as x = -2. The line y = x is diagonal with slope 1, and x = -2 is itself vertical; neither has the same direction as y = -2.

A point \((x,y)\) lies on the graph when \(y=f(x)=x^2\). For \((3,9)\), \(3^2=9\), so it satisfies the condition. In option B, \(3^2=9\), not 6; in option C, \((-3)^2=9\), not -9; and in option D, \(9^2=81\), not 3. Exam tip: substitute the x-coordinate into the function and compare the result with the y-coordinate.

For every real \(x\), \(x^2\ge 0\), so \(-x^2\le 0\) and hence \(f(x)=-x^2+1\le 1\). At \(x=0\), \(f(0)=1\), so the maximum value is 1. Option C is incorrect because the function equals 0, not -1, when \(x=\pm1\). Exam tip: a quadratic \(ax^2+bx+c\) with \(a<0\) opens downward, so its vertex gives the maximum value.

Replacing \(x\) by \(x-4\) in the parent function \(y=x^2\) shifts the graph horizontally 4 units to the right, so option A is correct. If the expression were \(x+4\), the graph would shift 4 units to the left; a constant added outside the square would cause a vertical shift. Exam tip: the graph of \(y=(x-h)^2\) is the graph of \(y=x^2\) shifted \(h\) units to the right.

In the vertex form \(y=(x-h)^2+k\), the vertex is \((h,k)\). Since \(x+3=x-(-3)\), we have \(h=-3\), and the outside constant gives \(k=2\). Therefore, the vertex is \((-3,2)\). Option B incorrectly changes the sign of the x-coordinate. Exam tip: reverse the sign of the number inside the bracket for the x-coordinate; the outside constant is the y-coordinate.

The expression \(|x-4|\) is smallest when the quantity inside the modulus is zero: \(x-4=0\). Thus, \(x=4\), and the minimum value of the function is \(0\). For instance, at \(x=1\), the value is \(3\), so it is not the minimum. Exam tip: \(|x-a|\) attains its minimum value, \(0\), at \(x=a\).

For transformations of y = |x|, a change inside the absolute-value symbol produces a horizontal shift. The expression |x + 5| can be written as |x - (-5)|, so the standard graph is translated 5 units to the left. This is also confirmed by the vertex: y = |x| has vertex (0, 0), while |x + 5| is zero when x = -5, giving vertex (-5, 0). Thus option A is correct. A positive number inside the brackets shifts left, not right, because the required input for the vertex becomes negative. The options describing upward or downward movement would require a constant outside the absolute value, such as |x| + 5 or |x| - 5.

In f(x)=2|x|, the y-value is doubled for every x compared with y=|x|. Therefore, the graph moves farther from the x-axis and is vertically stretched. It is not a horizontal shift because the expression inside the modulus has not changed. Exam tip: in y=a|x|, |a|>1 causes a vertical stretch, while 0<|a|<1 causes a vertical compression.

A square-root graph begins when the expression inside the radical is zero: \(x+4=0\), so \(x=-4\) and \(y=\sqrt{0}=0\). Therefore, the starting point is \((-4,0)\). Option B has the wrong sign for the x-coordinate. Exam tip: The starting point of \(y=\sqrt{x-a}\) is \((a,0)\).

The direct answer is D: \\((-1,1)\\) is not on the graph. For \\(y=\sqrt{x}\\), the input x must be non-negative, and the point must satisfy y equal to the principal square root of x. Check A: \\(\sqrt9=3\\), so \\( (9,3)\\) is on the graph. Check B: \\(\sqrt{16}=4\\), so \\( (16,4)\\) is on it. Check C: \\(\sqrt1=1\\), so \\( (1,1)\\) is on it. Check D: x=-1 is outside the real domain of the square-root function; \\(\sqrt{-1}\\) is not a real number, so \\((-1,1)\\) cannot lie on this real graph. D is correct. A, B and C each satisfy the equation exactly. Do not confuse \\(y=\sqrt{x}\\) with \\(y^2=x\\), which would also include negative y-values; the square-root function uses the non-negative principal value.