Graphs of standard functions

This expression is the sum of distances from (1) and (-1), whose minimum is (2). In distance-type modulus graphs, use the gap between fixed points.

Since (f(x-4)+1=(x-4)^2+1), the vertex is ((4,1)). Writing the transformed equation first reduces mistakes.

The graph of (\sqrt{x-2}) starts at (x=2), giving (y=3). The starting point is crucial for square-root graphs.

For large (|x|), (\frac{1}{x+2}) approaches (0), so (y=-1). The outside constant shifts the horizontal asymptote.

Here (f(-x)=-f(x)), so the graph is symmetric about the origin. Always compute (f(-x)) to test symmetry.

Check whether \(f(-x)=f(x)\). \(f(x)=x^4-2x^2\). Then \(f(-x)=(-x)^4-2(-x)^2=x^4-2x^2=f(x)\). Since \(f(-x)=f(x)\), the graph is even and is symmetric about the y-axis.

When (x\ge2), (|x+2|=x+2) and (|x-2|=x-2), so the value is (4). Decide the interval before removing modulus signs.

Reflection across the \(x\)-axis maps every point \((x,y)\) to \((x,-y)\). Thus the parabola \(y=x^2\), which opens upward, becomes \(y=-x^2\), which opens downward. Option B is the unchanged original graph, not its reflection. Exam tip: reflection in the \(x\)-axis changes the sign of \(y\); reflection in the \(y\)-axis changes the sign of \(x\).

From (y=\sqrt{9-x^2}), we get (x^2+y^2=9) with (y\ge0). The square root gives only the upper semicircle.

For every real value of \(x\), \(\sqrt{x^2}=|x|\), so the given graph is exactly the graph of \(y=|x|\). The graph \(y=x\) agrees only when \(x\ge 0\); for negative \(x\), \(|x|=-x\). Exam tip: simplify \(\sqrt{x^2}\) as \(|x|\), not simply as \(x\).

When \(x<0\), we have \(|x|=-x\). Hence, \(f(x)=\frac{-x}{x}=-1\), so the graph lies on the part of the horizontal line \(y=-1\) for which \(x<0\). The function is undefined at \(x=0\). Exam tip: determine the value of \(|x|\) separately for positive and negative values of \(x\).

For (x)-intercepts, (|x^2-4|=0), so (x^2-4=0). An absolute value is zero only when its inside is zero.

Since (f(|x|)=|x|^2=x^2), the graph does not change. If the original function is even, using (|x|) may keep the same graph.

Because of (|x|), (f(-x)=f(x)), so the graph is symmetric about the (y)-axis. Modulus often creates even symmetry.

Let (t=|x|). Then (t^2=t+2), giving (t=2), so (x=\pm2). Substituting (t=|x|) helps in modulus equations.

(\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1), but (x=1) is excluded. Do not forget the hole in such graphs.

To find the x-intercepts, set \(y=0\): \(|x|-2=0\), so \(|x|=2\) and \(x=\pm2\). Thus, the intercepts are \((-2,0)\) and \((2,0)\), and the distance between them is \(2-(-2)=4\) units. The value 2 represents the distance of each intercept from the origin, not the distance between the intercepts. Exam tip: Set \(y=0\) whenever you need to find x-intercepts.

As (x) goes left, (2^{x-1}) approaches (0), so the asymptote is (y=3). The outside (+3) shifts the exponential graph upward.

The direct answer is A, \(x>-5\). A logarithm is defined only when its input is strictly positive; zero and negative inputs are not allowed. Here the input is x+5, so solve \(x+5>0\). Subtract 5 from both sides to obtain \(x>-5\). Thus every real number greater than −5 belongs to the domain, while −5 itself does not. Option A is correct. Option B, \(x\ge-5\), incorrectly includes x=−5, but then the logarithm input is zero. Option C, \(x<5\), uses the wrong boundary and includes many values for which x+5 is nonpositive. Option D, \(x\ne-5\), excludes only −5 but still allows numbers less than −5, producing a logarithm of a negative number. Memory cue: for \(\log_b(g(x))\), write \(g(x)>0\), never \(g(x)\ge0\).

The log graph has an asymptote where its input approaches (0), so (x-2=0). In log graphs, the boundary of the domain gives the asymptote.

The maximum of (\sin x) is (1), attained at (x=\frac{\pi}{2}). Remember key points of standard trigonometric graphs.

The minimum of (\cos x) is (-1), which occurs at (x=\pi). When reading graphs, identify maximum and minimum points separately.

The direct answer is option A, \(\pi\). A period is the smallest positive horizontal distance after which a graph repeats exactly. For the tangent function, \(\tan(x+\pi)=\tan x\), because sine and cosine both change sign after adding \(\pi\), so their quotient stays unchanged. Therefore the graph repeats every \(\pi\), not merely after a larger multiple. Option A is correct. Option B, \(2\pi\), is also a repeating distance, but it is not the fundamental or smallest period, so it is less correct in a period MCQ. Option C, \(\frac{\pi}{2}\), is not a period; tangent values do not generally repeat after that shift. Option D, 1, is not a universal repeating distance. Also remember that tangent is undefined at \(x=\frac{\pi}{2}+k\pi\), which produces repeating branches. Exam cue: sine and cosine have period \(2\pi\), while tangent and cotangent have period \(\pi\).

Because (x-\frac{\pi}{4}) is inside, the graph shifts (\frac{\pi}{4}) to the right. Inner transformations move opposite to the sign.

It is the sum of distances from (-1) and (3), which stays minimum between them. The sum of distances to two points is minimized between those points.