Graphs of standard functions

The graph of a function is the set of all ordered pairs (x,f(x)). Here, f(1)=1^2=1, f(2)=2^2=4, and f(3)=3^2=9, so the graph is {(1,1),(2,4),(3,9)}. Option C represents the identity function y=x, while option B gives 6 instead of the correct value 9 for x=3. Exam tip: calculate f(x) for every element of the domain before forming the ordered pairs.

Using the rule \(y=x+2\), when \(x=1\), \(y=3\), and when \(x=3\), \(y=5\). Thus, the ordered pairs are \((1,3)\) and \((3,5)\), so option A is correct. Option C reverses the ordered-pair arrangement, while option D represents the identity rule \(y=x\). Exam tip: Always write graph points in the order \((x,y)\).

The graph of a function consists of ordered pairs of the form (x,f(x)). Here, f(0)=0, f(1)=3, and f(2)=6, so the graph is {(0,0),(1,3),(2,6)}. In option C, the components of the ordered pairs are reversed. Exam tip: evaluate the function at every domain element and write each result as (x,f(x)).

In (f(x)=x), (y=x) for every (x), so the graph is a straight line through the origin. In exams, test equal (x) and (y) values.

In (f(x)=c), the value of (y) remains (c) for all (x). So the graph is the horizontal line (y=c).

Since (x^2) is always (0) or positive, the parabola (y=x^2) opens upward. The point ((0,0)) is the vertex.

The direct answer is option B: the parabola opens downward. For y=-x^2, x^2 is always non-negative, so -x^2 is always zero or negative. The vertex is (0,0), and points such as (1,-1) and (-1,-1) lie below the x-axis. Therefore the two arms go downward. Option B is correct. Option A would describe y=x^2, where the coefficient of x^2 is positive. Options C and D describe sideways parabolas such as x=y^2 or x=-y^2, not a graph written as y in terms of x with a squared x. Memory cue: in y=ax^2, a>0 means upward and a<0 means downward.

Since (|-x|=|x|), (f(x)=|x|) is an even function and is symmetric about the (y)-axis. Replace (x) by (-x) to test symmetry.

The governing point condition for a function graph is that a point (x,y) lies on y=f(x) exactly when y=f(x). Test the coordinates in the options. For x=0, f(0)=0³=0, so (0,0) lies on the graph and option C is correct. For (0,1), the output would need to be 1, but f(0)=0. For (1,0), the function gives f(1)=1, not 0. For x=−1, f(−1)=(−1)³=−1, so the corresponding graph point is (−1,−1), not (−1,1). Therefore the cubic graph passes through the origin. This conclusion follows directly by substitution and does not require plotting the entire curve. The odd symmetry of x³ also agrees with the fact that the origin is on the graph.

In real numbers, (\sqrt{x}) is defined only when (x\ge 0). So its graph starts on the right side.

The direct answer is C: x=0. A function containing 1/x is defined only when its denominator is non-zero. Therefore we require x≠0. At x=0, the expression becomes 1/0, which has no real value, so the point is not part of the graph. Option A, x=1, is allowed because 1/1=1. Option B, x=-1, is allowed because 1/(-1)=-1. Option C is correct because x=0 makes the denominator zero. Option D, x=2, is allowed because 1/2 is a valid real number. The domain is all real numbers except zero, written as ℝ\{0}, and the graph has two branches with x=0 as a vertical asymptote. Always inspect denominators before accepting a value. Do not confuse an excluded input with an output: y=0 is also never obtained, but the question asks which x-value is forbidden. Exam cue: denominator zero means undefined, so remove that input from the domain.

The governing concept is the range and minimum of the square function. For every real number x, the square x^2 is non-negative, so x^2 ≥ 0. Equality is achieved when x=0, because 0^2=0. Therefore the least, or minimum, value of f(x)=x^2 is 0, making option A correct. The value 1 is attained at x=1 and x=-1, but it is not the least value because values between 0 and 1 are also possible. The value -1 cannot occur because a real square is never negative. The function does have a minimum, located at the vertex (0,0) of its upward-opening parabola.

The governing concept is the absolute-value function and its graph. For every real number x, |x| denotes the distance of x from zero, so it can never be negative: |x|≥0. The least possible distance is 0, and this occurs only when x=0. Therefore f(0)=|0|=0, so the graph has its vertex and unique minimum point at (0,0); option A is correct. The points (1,1) and (−1,1) lie on the V-shaped graph, but their function value is 1, which is larger than 0. The point (0,1) is not on the graph because substituting x=0 gives y=0, not y=1. Thus the minimum value is 0 and it is attained at the point (0,0).

The direct answer is A: the point lies on the graph because \\(4=2^2\\). A point \\((x,y)\\) is on the graph \\(y=x^2\\) exactly when its coordinates satisfy that equation. Substitute x=2: the right side is \\(2^2=4\\), which equals the given y-coordinate. Therefore \\((2,4)\\) is on the graph. A is correct. Option B says \\(2=4^2\\), but \\(4^2=16\\), not 2, and it reverses the required substitution. Option C says \\(4=2+2\\), which is numerically true but is not the graph equation \\(y=x^2\\); a true unrelated statement does not prove graph membership. Option D says \\(2=4-2\\), also numerically true but irrelevant to the equation. The reliable method is always to substitute both coordinates into the actual equation, not merely find any true relation between them.

On (y=x), both coordinates are equal. In ((4,5)), (4\neq 5), so this point is not on the graph.

\\(f(-3)=(-3)^2=9\\) and \\(f(3)=3^2=9\\), so the two values are equal. Option D is incorrect because the negative of \\(f(3)=9\\) is \\(-9\\), whereas \\(f(-3)=9\\). Exam tip: \\(f(x)=x^2\\) is an even function, so \\(f(-x)=f(x)\\), and its graph is symmetric about the y-axis.

\(f(-1)=(-1)^3=-1\) and \(f(1)=1^3=1\). Therefore, \(f(-1)=-f(1)\). Option A is incorrect because the two values are not equal, and option C is incorrect because \(-1<1\). Exam tip: a function satisfying \(f(-x)=-f(x)\) is odd, and \(x^3\) is an odd function.

In the equation \(y=3\), the function value remains 3 for every value of \(x\), so its graph represents a constant function. It is a horizontal line parallel to the \(x\)-axis. An identity function has \(y=x\), so its graph is a slanting line instead. Exam tip: A function whose output does not change when \(x\) changes is a constant function.

(f(x)=2x) is a linear function and at (x=0), (y=0), so the line passes through the origin. The coefficient gives the slope.

Every point on the \(y\)-axis has \(x=0\). Therefore, \(f(0)=0+2=2\), so the \(y\)-coordinate of the \(y\)-intercept is 2 and the intercept point is \((0,2)\). The value \(−2\) would result from a function such as \(x−2\), not \(x+2\). Exam tip: To find a graph’s \(y\)-intercept, substitute \(x=0\).

At the (x)-intercept, (y=0), so (x-3=0) gives (x=3). For intercepts, set the proper coordinate to zero.

The governing concept is horizontal translation of the basic absolute-value graph y=|x|. An absolute-value expression is smallest when its inside quantity is zero. Set x-2=0, which gives x=2. At that x-value, f(2)=|2-2|=0, so the vertex, or minimum point, is (2,0). Therefore option B is correct. The graph y=|x-2| is the graph of y=|x| shifted two units to the right. The point (0,0) belongs to the unshifted graph, not this one; (-2,0) would correspond to |x+2|; and (0,2) confuses the horizontal shift with a vertical value. Checking the zero inside the modulus avoids these errors.

For every value of \(x\), the function \(f(x)=x^2+1\) gives a \(y\)-value that is 1 greater than that of \(y=x^2\). Thus, the parabola keeps the same shape and horizontal position but shifts upward by 1 unit. Exam tip: in \(y=g(x)+k\), a positive \(k\) shifts the graph upward by \(k\) units.

The minimum of ((x-1)^2) is (0) when (x=1), so the vertex is ((1,0)). Set the inside of the bracket to zero.

In (\sqrt{x}), the smallest allowed (x)-value is (0), and (\sqrt{0}=0). So the graph starts from ((0,0)).