Graphing Trig Functions — Free Trigonometry Review Games.

Amplitude = |4| = 4.

The standard sine function completes one cycle in 2*pi.

The vertical shift moves the midline to y = 3.

Period = 2*pi/|b| = 2*pi/2 = pi.

The negative sign reflects the graph over the x-axis.

Period of tan = pi/|b| = pi/2.

Amplitude = 3. Period = 2*pi/pi = 2.

cos(x - pi/2) = sin(x) by the co-function identity.

y = sin(x + c) shifts left by c. Phase shift is pi/4 left.

Period = 2*pi/3. In interval 2*pi, there are 2*pi / (2*pi/3) = 3 periods.

Period = 2*pi/b = 4*pi means b = 1/2. So y = 2*sin(x/2).

-2*cos(x) ranges from -2 to 2. Adding 5: [3, 7].

A=3, period=2*pi/2=pi, phase shift=pi/6 right, midline y=1.

tan(x) is undefined at x = pi/2 + n*pi, creating vertical asymptotes.

csc(x) = 1/sin(x), so it has the same period as sin(x): 2*pi.

Amplitude is defined as the absolute value of the coefficient in front of the trig function, so \(|-5| = 5\). The distractor \(-5\) is wrong because amplitude represents a distance from the midline and cannot be negative. Always take the absolute value of the leading coefficient to find amplitude, regardless of its sign.

The period formula \(\frac{2\pi}{|b|}\) with \(b = \frac{1}{3}\) gives \(\frac{2\pi}{1/3} = 6\pi\). The distractor \(\frac{2\pi}{3}\) incorrectly multiplies instead of dividing by \(b\). Remember that dividing \(x\) by a number stretches the graph horizontally, increasing the period.

The midline is the horizontal line the graph oscillates around, determined by the vertical shift \(d\) in \(y = \cos(x) + d\), which here is \(-2\). The distractor \(y = 2\) has the wrong sign, confusing an upward and downward shift. Always identify the constant added or subtracted outside the trig function to locate the midline.

Cosine is an even function, meaning \(\cos(-x) = \cos(x)\) for all \(x\), so the graphs are identical. The distractor \(y = -\cos(x)\) describes a reflection over the x-axis, which is a different transformation than reflecting over the y-axis for an even function. Knowing that cosine is even and sine is odd helps quickly simplify negative-angle expressions.

Tangent is undefined wherever cosine equals zero, which occurs at \(x = \frac{\pi}{2} + k\pi\) for any integer \(k\), creating vertical asymptotes there. The distractor \(x = k\pi\) is wrong because those are the x-intercepts of tangent, not the excluded values. Since \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), always check where the denominator is zero to find domain restrictions.

The tangent function repeats every \(\pi\) radians because its pattern between consecutive asymptotes spans exactly \(\pi\). The distractor \(2\pi\) is the period of sine and cosine, not tangent, so it is a common confusion. Remember that tangent's period is half that of sine and cosine due to its different graph structure.

Sine equals zero at \(x = 0\), \(x = \pi\), and \(x = 2\pi\) within this closed interval, giving three intercepts. The distractor 2 would be correct only on a half-open interval that excludes one endpoint. Always check whether the interval endpoints are included when counting zeros of a periodic function.

Plugging in \(x = 0\) gives \(\cos(0) = 1\), which is the value where the graph crosses the y-axis. The distractor \(0\) is actually the y-intercept of sine, not cosine, a common mix-up between the two functions. Remembering that cosine starts at its maximum value while sine starts at zero helps distinguish their graphs quickly.

Subtracting a constant outside the function shifts the entire graph vertically downward by that amount, so \(-4\) moves it down 4 units. The distractor "Shifting up 4 units" reverses the correct direction, a frequent sign error. Always treat the sign of the outside constant as the direct indicator of vertical shift direction.

Since \(\sec(x) = \frac{1}{\cos(x)}\), it inherits the same period as cosine, which is \(2\pi\). The distractor \(\pi\) incorrectly assumes secant behaves like tangent rather than cosine. Reciprocal trig functions always share the period of their corresponding base function.

The amplitude equals the absolute value of the coefficient multiplying the trig function, here \(\left|\frac{1}{2}\right| = \frac{1}{2}\). The distractor \(2\) confuses the coefficient with its reciprocal. Always read the amplitude directly from the number multiplying sine or cosine, taking its absolute value.

Tangent has no maximum or minimum value since it extends to infinity between asymptotes, so amplitude is undefined for it. The distractor \(y = 3\sin(x)\) clearly has a bounded range and thus a well-defined amplitude of 3. Amplitude only applies to functions with a bounded, oscillating range like sine, cosine, and their reciprocal-free transformations.

Using the period formula \(\frac{2\pi}{|b|}\) with \(b = \frac{1}{4}\) gives \(\frac{2\pi}{1/4} = 8\pi\). The distractor \(4\pi\) mistakenly uses \(b = \frac{1}{2}\) instead of \(\frac{1}{4}\). A smaller value of \(b\) stretches the graph and produces a longer period.

Rewriting as \(\cos\left(2\left(x - \frac{\pi}{6}\right)\right)\) shows the phase shift is \(\frac{c}{b} = \frac{\pi/3}{2} = \frac{\pi}{6}\) to the right. The distractor \(\frac{\pi}{3}\) right forgets to divide by the horizontal compression factor \(b = 2\). Always factor out \(b\) before reading off the phase shift value.

Rewriting as \(3\sin\left(2\left(x + \frac{\pi}{2}\right)\right)\) gives period \(\frac{2\pi}{2} = \pi\) and phase shift \(\frac{\pi}{2}\) left since the shift is \(-\frac{c}{b}\). The distractor "Period \(2\pi\)" ignores the factor of 2 inside the sine function that compresses the graph. Both period and phase shift require factoring out the coefficient of \(x\) first.

Tangent has vertical asymptotes at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) within this interval, giving exactly two. The distractor 1 misses one of the two asymptotes that occur every \(\pi\) units starting from \(\frac{\pi}{2}\). Counting asymptotes requires listing all values of \(\frac{\pi}{2} + k\pi\) that fall inside the given interval.

Setting \(\sin\left(x - \frac{\pi}{2}\right) = 0\) requires \(x - \frac{\pi}{2} = k\pi\), so \(x = \frac{\pi}{2} + k\pi\), giving \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) in the interval. The distractor \(x = 0, \pi, 2\pi\) lists the intercepts of the untranslated \(\sin(x)\), ignoring the phase shift. Always substitute the entire shifted argument into the zero condition rather than using the base function's intercepts.

Since period \(= \frac{2\pi}{b} = \pi\), solving gives \(b = 2\), and amplitude 3 gives the coefficient 3, matching \(y = 3\cos(2x)\). The distractor \(y = 3\cos\left(\frac{x}{2}\right)\) has period \(4\pi\), which is too long. Always solve the period equation for \(b\) before assembling the final function.

The period of \(y = \cos(4x)\) is \(\frac{2\pi}{4} = \frac{\pi}{2}\), and dividing the interval length \(\pi\) by \(\frac{\pi}{2}\) gives 2 complete periods. The distractor 4 mistakes the coefficient itself for the number of periods without dividing correctly. Always divide the interval length by the actual period to count repetitions.

Adding a positive constant outside the sine function moves every point on the graph upward by that constant, so the graph shifts up 2 units. The distractor "Stretched vertically by 2" describes multiplying the function, not adding to it, which changes amplitude rather than position. Additive constants outside the trig function always control vertical position, not shape.

Subtracting inside the argument, as in \(x - \frac{\pi}{3}\), shifts the graph to the right by \(\frac{\pi}{3}\) because the function reaches the same output values at larger \(x\) values. The distractor "Left \(\frac{\pi}{3}\)" reverses the correct direction, a very common sign mistake. Remember that inside-the-function shifts behave oppositely to what the sign initially suggests.

The maximum of sine is 1, so \(4(1) - 1 = 3\) gives the maximum value of the transformed function. The distractor \(4\) forgets to subtract the vertical shift after applying the amplitude. Always compute amplitude times the sine or cosine extreme value first, then apply the vertical shift.

Since the coefficient is negative, the minimum occurs when \(\cos(x) = 1\), giving \(-3(1) + 2 = -1\). The distractor \(-3\) forgets to add the vertical shift of 2 after applying the amplitude and reflection. Negative amplitude coefficients flip which cosine extreme produces the maximum versus minimum value.

The period of tangent is \(\frac{\pi}{|b|}\), and with \(b = \frac{1}{2}\) this becomes \(\frac{\pi}{1/2} = 2\pi\). The distractor \(\pi\) is the period of the untransformed tangent function and ignores the horizontal stretch caused by \(b = \frac{1}{2}\). Tangent's period formula uses \(\pi\) in the numerator, unlike sine and cosine which use \(2\pi\).

Tangent's asymptotes occur where the argument equals \(\frac{\pi}{2} + k\pi\), so solving \(x - \frac{\pi}{4} = \frac{\pi}{2}\) gives \(x = \frac{3\pi}{4}\) as the nearest positive asymptote. The distractor \(x = \frac{\pi}{4}\) is actually where the graph crosses zero, not where it is undefined. Always set the shifted argument equal to the base function's asymptote condition and solve for \(x\).

Using the identity \(\cos(x + \pi) = -\cos(x)\), the graph is a reflection of cosine over the x-axis. The distractor \(y = \cos(x)\) ignores the phase shift of \(\pi\), which flips the entire curve. A horizontal shift of exactly half a period always produces the same effect as a reflection for sine and cosine graphs.

Multiplying \(x\) by 3 inside the function compresses the graph horizontally, completing three times as many cycles in the same interval, so the period shrinks by a factor of 3. The distractor "stretched horizontally" describes the opposite effect that would occur if \(b\) were a fraction rather than 3. Coefficients greater than 1 inside the argument always compress, while coefficients between 0 and 1 stretch.

Since \(\sec(x) = \frac{1}{\cos(x)}\), the function is undefined wherever \(\cos(x) = 0\), which happens at \(x = \frac{\pi}{2}\). The distractor referencing \(\sin\left(\frac{\pi}{2}\right) = 0\) is false since \(\sin\left(\frac{\pi}{2}\right) = 1\), and even if true it would relate to cosecant, not secant. Reciprocal trig functions are undefined exactly where their corresponding base function equals zero.

Using the period formula \(\frac{2\pi}{|b|}\) with \(b = \frac{\pi}{2}\) gives \(\frac{2\pi}{\pi/2} = 4\). The distractor \(2\pi\) incorrectly ignores the coefficient \(\frac{\pi}{2}\) inside the argument. When \(b\) contains \(\pi\), it often cancels with the \(2\pi\) in the numerator to give a rational period.

Factoring gives \(4\sin\left(2\left(x + \frac{\pi}{4}\right)\right)\), so the phase shift is \(\frac{\pi}{4}\) to the left since the shift equals \(-\frac{c}{b}\). The distractor \(\frac{\pi}{2}\) left skips the division by the horizontal compression factor \(b = 2\). Factoring out the coefficient of \(x\) before identifying the shift prevents this common error.

Cotangent, like tangent, repeats every \(\pi\) radians because it is the reciprocal of tangent and shares the same fundamental period. The distractor \(2\pi\) mistakenly applies the sine and cosine period instead of the tangent-family period. Both tangent and cotangent have a period of \(\pi\), which is shorter than the \(2\pi\) period of sine, cosine, secant, and cosecant.

Since period \(\pi\) requires \(b = 2\), and the shift is written as \(x - \frac{\pi}{4}\) inside the compressed argument, the correct form is \(3\cos\left(2\left(x - \frac{\pi}{4}\right)\right)\), which expands to \(3\cos\left(2x - \frac{\pi}{2}\right)\). The distractor \(3\cos\left(2x - \frac{\pi}{4}\right)\) fails to multiply the shift by \(b\) when expanding, which actually produces a phase shift of only \(\frac{\pi}{8}\). Always build the equation in factored form \(b(x - h)\) first to avoid shift errors when a horizontal compression is present.

The amplitude of 3 stretches sine's range to \([-3, 3]\), and adding 2 shifts that entire interval up to \([-1, 5]\), while the phase shift does not affect the range at all. The distractor \([-3, 3]\) forgets to apply the vertical shift after establishing the amplitude-stretched range. Phase shifts only move a graph horizontally and never change its amplitude or range.

Since the period of \(y = \tan(2x)\) is \(\frac{\pi}{2}\), asymptotes recur every \(\frac{\pi}{2}\) units, giving four asymptotes at \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\) within $[0, 2\pi)$. The distractor 2 assumes the untransformed tangent period of \(\pi\), missing the effect of the horizontal compression from the coefficient 2. Horizontal compression of tangent doubles the number of asymptotes in a fixed interval compared to the base function.

Cosine reaches its maximum at \(x = 0\) shifted by \(\frac{\pi}{2}\) inherently — actually \(\cos(x)\) itself has max at \(x=0\), but checking values, \(\cos\left(\frac{\pi}{2}\right)=0\), so testing \(y=\sin(x)\): it reaches max 1 at \(x=\frac{\pi}{2}\) and min \(-1\) at \(x=\frac{3\pi}{2}\), matching the description, meaning the correct choice is actually \(\sin(x)\), so the mechanism is that sine attains its maximum a quarter period after zero and its minimum three-quarters through the cycle. The distractor \(y = \cos(x)\) is wrong because \(\cos\left(\frac{\pi}{2}\right) = 0\), not a maximum, so its extreme points occur at different x-values. Identifying key maximum and minimum locations relative to a period is the fastest way to distinguish sine from cosine graphs.

The negative coefficient causes a reflection, the amplitude is \(|-2| = 2\), the period is \(\frac{2\pi}{3}\) from \(b = 3\), the shift is left \(\frac{\pi}{4}\) from the factored form \(x + \frac{\pi}{4}\), and the \(-1\) shifts the midline down 1 unit. The distractor with "shift right \(\frac{\pi}{4}\)" incorrectly reads the sign inside the parentheses, since \(x + \frac{\pi}{4}\) actually indicates a leftward shift. When multiple transformations combine, each parameter — amplitude, period, phase shift, vertical shift, and reflection — must be identified separately from the equation's factored form.

Since the tangent period formula is \(\frac{\pi}{|b|}\), setting \(\frac{\pi}{2} = \frac{\pi}{b}\) gives \(b = 2\), and writing the shift in factored form as \(x - \frac{\pi}{8}\) correctly represents a rightward shift. The distractor \(y = \tan\left(2x - \frac{\pi}{8}\right)\) fails to factor out \(b\), so expanding the correct equation actually gives \(2x - \frac{\pi}{4}\), not \(2x - \frac{\pi}{8}\). Building tangent equations from period and phase shift requires solving for \(b\) using \(\frac{\pi}{b}\) rather than \(\frac{2\pi}{b}\) used for sine and cosine.

Factoring gives \(\sin\left(2\left(x - \frac{\pi}{2}\right)\right)\), showing the true phase shift is \(\frac{\pi}{2}\) to the right, not \(\pi\) as the unfactored form might suggest. The distractor \(\pi\) right comes from mistakenly reading the constant term directly without dividing by the coefficient \(b = 2\). Phase shift must always be calculated after factoring out the coefficient of \(x\), never read directly from the unfactored equation.

Setting \(2x - \frac{\pi}{3} = \frac{\pi}{2} + k\pi\) and solving for \(x\) gives \(x = \frac{5\pi}{12} + \frac{k\pi}{2}\), which yields \(\frac{5\pi}{12}\) and \(\frac{11\pi}{12}\) within \([0, \pi]\). The distractor \(x = \frac{\pi}{6}, \frac{2\pi}{3}\) comes from incorrectly setting the argument equal to \(k\pi\) instead of \(\frac{\pi}{2} + k\pi\), which is the zero condition for cosine rather than sine. Solving multi-step trig equations requires using the correct zero condition for the specific function before isolating \(x\).

Substituting gives \(3\sin\left(\frac{\pi}{2}\right) + 1 = 3(1) + 1 = 4\), since \(\sin\left(\frac{\pi}{2}\right) = 1\). The distractor \(1\) mistakenly assumes \(\sin\left(\frac{\pi}{2}\right) = 0\), confusing it with \(\sin(0)\) or \(\sin(\pi)\). Careful substitution of the doubled angle before evaluating the sine value is essential to avoid arithmetic errors.

Solving \(\frac{2\pi}{b} = \frac{2\pi}{3}\) gives \(b = 3\), and the reflection combined with amplitude 4 produces the negative coefficient, so \(y = -4\cos(3x)\) satisfies all given features. The distractor \(y = 4\cos(3x)\) has the correct period and amplitude but omits the reflection over the x-axis. Always translate each stated feature — amplitude, period, and reflection — into a distinct part of the equation before combining them.

Since \(\tan(-\theta) = -\tan(\theta)\), replacing \(b\) with \(-b\) reflects the graph over the y-axis, but because tangent is an odd function this visually appears identical to reflecting over the x-axis. The distractor "changes the amplitude" is incorrect because tangent has no defined amplitude to alter. Recognizing that tangent is an odd function explains why sign changes in \(b\) produce a mirrored graph rather than a shifted or stretched one.

Setting \(x - \frac{\pi}{2} = \frac{\pi}{2} + k\pi\) and solving gives \(x = \pi + k\pi\), producing \(x = 0, \pi, 2\pi\) within the closed interval since \(k=-1,0,1\) all land in range. The distractor \(x = \frac{\pi}{2}, \frac{3\pi}{2}\) mistakenly uses the unshifted tangent asymptote locations without accounting for the horizontal translation. Always substitute the full shifted argument into the asymptote condition rather than reusing the base function's asymptote positions.

Amplitude equals half the difference between max and min, \(\frac{7-(-1)}{2} = 4\), and vertical shift equals the average of max and min, \(\frac{7+(-1)}{2} = 3\). The distractor "Amplitude 8" mistakenly uses the full difference between max and min instead of halving it. Always halve the max-minus-min difference for amplitude and average the max and min for the midline shift.

A horizontal shift of exactly half the period, \(\pi\) for sine, produces \(\sin(x+\pi) = -\sin(x)\), meaning the shifted graph equals the reflected graph, so the pair \(y=\sin(x)\) and \(y=-\sin(x)\) represents this identical outcome. The distractor \(y = \tan(x)\) and \(y = \tan(x+\pi)\) is wrong because tangent's period is already \(\pi\), so shifting by \(\pi\) produces the exact same graph rather than a reflection. Recognizing that a half-period shift equals a reflection for sine and cosine, but a full-period repeat for tangent, prevents confusion between these two distinct graph behaviors.