Graphs of standard functions

On the (x)-axis \(y=0\) so \(2x-5=0\) gives \(x=\frac{5}{2}\). For an intercept set the other coordinate equal to zero.

\( |x-3| \) is minimum (0) when \(x=3\). For a modulus graph set the inside expression equal to zero.

The greatest integer function gives the largest integer less than or equal to (x). Hence (\lfloor 2.7 \rfloor=2).

For the greatest integer function \(f(x)=\lfloor x\rfloor\), values just to the left of \(x=3\) approach 2, whereas \(f(3)=3\) and values just to the right are 3. Thus the left-hand and right-hand limits are different, so there is a jump discontinuity at \(x=3\). The function is defined for every real \(x\), making option D incorrect. Exam tip: The graph of \(\lfloor x\rfloor\) has a jump discontinuity at every integer.

For the signum function, \(\operatorname{sgn}(x)=-1\) when \(x<0\), \(0\) when \(x=0\), and \(1\) when \(x>0\). Since \(-8<0\), we get \(f(-8)=-1\). The value 0 would apply only at \(x=0\). Exam tip: For a signum function, check the sign of the input rather than its magnitude.

The function \(\operatorname{sgn}(x-2)\) equals \(-1\) when \(x-2<0\) and \(1\) when \(x-2>0\). Its value changes when \(x-2=0\), giving \(x=2\); the unequal one-sided limits therefore produce a jump discontinuity there. \(x=0\) would be the change point for \(\operatorname{sgn}(x)\), not for this shifted function. Exam tip: for \(\operatorname{sgn}(g(x))\), first solve \(g(x)=0\) to locate possible jumps.

(y=x^3) is an odd function because (f(-x)=-f(x)). The graph of an odd function is symmetric about the origin.

Completing the square gives \(y=x^2-6x+11=(x-3)^2+2\). In the standard form \(y=(x-h)^2+k\), the vertex is \((h,k)\); hence it is \((3,2)\). Option B has the wrong sign for the x-coordinate, while option C results from incorrectly reading the coefficients as coordinates. Exam tip: rewrite a quadratic in the form \(y=a(x-h)^2+k\) to identify its vertex directly.

The expression represents the sum of the distances of x from −1 and 3 on the number line. For −1≤x≤3, this sum remains constant at 3−(−1)=4, so the minimum value is 4. It cannot be 0 because both absolute-value terms cannot be zero simultaneously. Exam tip: the minimum of |x−a|+|x−b| is |a−b|.

At any point on the x-axis, \(y=0\). Therefore, \(|2x-6|=0\), and an absolute value is zero only when its inside expression is zero: \(2x-6=0\). Thus \(x=3\), giving the intercept \((3,0)\). Option D is incorrect because for \(x=6\), \(y=|12-6|=6\), not zero. Exam tip: To find an x-intercept, set \(y=0\).

In the function \(y=5\), the value of \(y\) remains fixed at 5 for every \(x\). Thus, all points are of the form \((x,5)\), giving a horizontal line parallel to the \(x\)-axis. It does not pass through the origin because when \(x=0\), \(y=5\). Exam tip: the graph of \(y=c\) is always a line parallel to the \(x\)-axis.

For the identity function \(f(x)=x\), every point \((x,y)\) on the graph must satisfy \(y=x\). In option A, both coordinates of \((-4,-4)\) are equal, so the point lies on the graph. In the other options, the x- and y-coordinates are unequal. Exam tip: The graph of the identity function is the straight line \(y=x\) passing through the origin.

At an intersection, both functions have the same \(y\)-value, so set \(x^2=2x+3\). This gives \(x^2-2x-3=0\), or \((x-3)(x+1)=0\). Hence \(x=3\) or \(x=-1\), making option A correct. Exam tip: To find intersection coordinates, first equate the two function expressions and solve for \(x\).

At an intersection, the y-values are equal, so x² = |x|. The value x = 0 is a solution. For x ≠ 0, divide by |x| to get |x| = 1, which gives x = −1 or x = 1. Therefore, the intersection x-values are −1, 0, and 1.

When \(x>0\), \(|x|=x\). Hence, \(f(x)=\frac{x}{|x|}=\frac{x}{x}=1\), so the graph lies on the horizontal line \(y=1\) for all positive values of \(x\). The line \(y=-1\) applies when \(x<0\), not in the given interval. Exam tip: \(\frac{x}{|x|}\) gives the sign of \(x\), with values \(1\) for positive \(x\) and \(-1\) for negative \(x\).

When \(x>-1\), the value of \(x\) is greater than \(-1\). Therefore, the smaller of \(x\) and \(-1\) is always \(-1\), so \(y=\min(x,-1)=-1\). Option B is incorrect because \(y=x\) applies when \(x\leq -1\). Exam tip: For a minimum function, select the smaller expression and check the boundary value \(x=-1\).

Here, \(f(x)=x^4\). Thus, \(f(-x)=(-x)^4=x^4=f(x)\), so the function is even and its graph is symmetric about the \(y\)-axis. Option B describes the symmetry of an odd function, but \(x^4\) does not satisfy \(f(-x)=-f(x)\). Exam tip: an even function has \(y\)-axis symmetry, whereas an odd function has symmetry about the origin.

Since |x| ≥ 0 for every real x, y = −|x| ≤ 0. The value 0 occurs at x = 0, and the outputs decrease without bound as |x| grows. Thus the range is (−∞, 0].

For \(f(x)=\frac{1}{x^2}\), we have \(f(-x)=\frac{1}{(-x)^2}=\frac{1}{x^2}=f(x)\). Therefore, the function is even and its graph is symmetric about the y-axis. Option B is incorrect because the function is undefined at \(x=0\), so the graph does not pass through the origin. Its domain is \(\mathbb{R}\setminus\{0\}\) and its range is \((0,\infty)\), not \(\mathbb{R}\). Exam tip: If \(f(-x)=f(x)\), the graph has symmetry about the y-axis.

The domain of \(f(x)=\frac{1}{x}\) is \(x\ne 0\). Since \(f(-x)=-\frac{1}{x}=-f(x)\), the function is odd and its graph is symmetric about the origin. Option B is incorrect because symmetry about the \(y\)-axis requires an even function. The function is undefined at \(x=0\), and its graph lies in both the first and third quadrants. Exam tip: if \(f(-x)=-f(x)\), the graph is symmetric about the origin.

Every point on the \\(x\\)-axis has \\(y=0\\). Thus, \\(0=x^2-4\\), so \\(x^2=4\\) and \\(x=\\pm2\\). Therefore, the intersection points are \\((-2,0)\\) and \\((2,0)\\). The points in option B have \\(x=0\\), so they lie on the \\(y\\)-axis instead. Exam tip: To find intersections with the \\(x\\)-axis, set \\(y=0\\).

Replacing (x) by (x-5) shifts the graph (5) units right. In horizontal shift read the sign oppositely.

The graph of y = f(x + 3) is obtained by shifting the graph of y = f(x) 3 units to the left. Here f(x) = x^2, so the vertex of y = (x + 3)^2 is (-3, 0). A rightward shift is incorrect because a positive number added inside the function causes a horizontal shift to the left. Exam tip: f(x - a) shifts a graph a units right, while f(x + a) shifts it a units left.

In \(y=x^2+6\), 6 is added outside the function, so for every \(x\), the new \(y\)-value is 6 greater than before. Thus the entire parabola shifts 6 units upward. A downward shift would correspond to \(y=x^2-6\). Exam tip: in \(y=f(x)+k\), a positive \(k\) shifts the graph upward by \(k\) units.

The direct answer is option A: reflection in the x-axis. Start with a point \((a,b)\) on \(y=\sqrt{x}\), where \(b=\sqrt a\) and \(a\ge0\). In \(y=-\sqrt{x}\), the same x-value gives height \(-b\), so the point becomes \((a,-b)\). This is exactly reflection across the x-axis. Option A is correct. Option B would replace x by \(-x\), producing a y-axis reflection such as \(y=\sqrt{-x}\), not the given function. Option C would require adding 1 outside the function, such as \(y=\sqrt{x}+1\). Option D would require replacing x by \(x-1\), such as \(y=\sqrt{x-1}\). The domain remains \(x\ge0\), while the range changes from \(y\ge0\) to \(y\le0\). Exam cue: a minus outside a function changes y-values; a minus inside the input changes x-direction.