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Algebra - Cubic, trigonometric, and logarithmic functions and their asymptotes-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Cubic functions follow the general form f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d. The leading coefficient aa determines the end behavior: if a>0a > 0, the graph starts in the third quadrant and ends in the first; if a<0a < 0, it starts in the second and ends in the fourth. The graph can have up to two turning points and between one and three xx-intercepts.

Graph of a cubic function y = x^3 - 4x showing three x-intercepts and two turning points.
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The logarithmic function f(x)=log⁡b(x−h)+kf(x) = \log_b(x - h) + k is the inverse of the exponential function. It possesses a vertical asymptote at x=hx = h, which defines the boundary of the domain (x>hx > h). The function is only defined for positive arguments.

Graph of log base 2 of (x-1) showing a vertical asymptote at x=1.
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Trigonometric functions like y=tan⁡(x)y = \tan(x) exhibit periodic behavior with vertical asymptotes where the function is undefined (i.e., where cos⁡(x)=0\cos(x) = 0). For y=tan⁡(x)y = \tan(x), these occur at x=90∘+180∘nx = 90^\circ + 180^\circ n. The period of the tangent function is 180∘180^\circ or π\pi radians.

Graph of y = tan(x) between -180 and 180 degrees showing vertical asymptotes at +/- 90 degrees.
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The yy-intercept of any function f(x)f(x) is found by evaluating f(0)f(0). For cubic functions ax3+bx2+cx+dax^3+bx^2+cx+d, the yy-intercept is always dd. For logarithmic functions, a yy-intercept only exists if the domain includes x=0x=0.

📐Formulae

f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d

y=log⁡b(x)  ⟺  by=xy = \log_b(x) \iff b^y = x

Asymptotes of tan⁡(x):x=90∘+180∘n, where n∈Z\text{Asymptotes of } \tan(x): x = 90^\circ + 180^\circ n, \text{ where } n \in \mathbb{Z}

Change of base: log⁡a(b)=log⁡c(b)log⁡c(a)\text{Change of base: } \log_a(b) = \frac{\log_c(b)}{\log_c(a)}

💡Examples

Problem 1:

Find the xx-intercepts and the general shape of the cubic function y=x3−4xy = x^3 - 4x.

Solution:

To find the xx-intercepts, set y=0y = 0: x3−4x=0x^3 - 4x = 0 x(x2−4)=0x(x^2 - 4) = 0 x(x−2)(x+2)=0x(x - 2)(x + 2) = 0 Thus, x=0,x=2,x=−2x = 0, x = 2, x = -2.

Explanation:

The graph crosses the xx-axis at three points: (−2,0)(-2, 0), (0,0)(0, 0), and (2,0)(2, 0). Since the leading coefficient a=1a=1 is positive, the graph starts from the bottom-left and ends at the top-right.

Problem 2:

Determine the vertical asymptote and domain of the function f(x)=log⁡2(x+3)−5f(x) = \log_{2}(x + 3) - 5.

Solution:

The argument of the logarithm must be strictly greater than zero: x+3>0  ⟹  x>−3x + 3 > 0 \implies x > -3 The vertical asymptote occurs when the argument is zero: x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3

Explanation:

The domain of the function is x∈(−3,∞)x \in (-3, \infty) and the vertical asymptote is the line x=−3x = -3.

Problem 3:

Identify the first two positive vertical asymptotes for the function y=tan⁡(x)y = \tan(x) in degrees.

Solution:

The function y=tan⁡(x)y = \tan(x) is defined as sin⁡(x)cos⁡(x)\frac{\sin(x)}{\cos(x)}. Asymptotes occur where cos⁡(x)=0\cos(x) = 0. In the interval x>0x > 0, cos⁡(x)=0\cos(x) = 0 at: x=90∘,270∘,450∘…x = 90^\circ, 270^\circ, 450^\circ \dots

Explanation:

The first two positive vertical asymptotes are the lines x=90∘x = 90^\circ and x=270∘x = 270^\circ.

Problem 4:

Sketch the function f(x)=−(x+2)(x−1)(x−3)f(x) = -(x+2)(x-1)(x-3) and identify the yy-intercept.

Cubic graph with x-intercepts at -2, 1, 3 and y-intercept at -6.

Solution:

  1. Identify the xx-intercepts by setting f(x)=0f(x) = 0: x=−2,1,3x = -2, 1, 3.
  2. Find the yy-intercept by setting x=0x = 0: f(0)=−(0+2)(0−1)(0−3)=−(2)(−1)(−3)=−6f(0) = -(0+2)(0-1)(0-3) = -(2)(-1)(-3) = -6.
  3. Determine end behavior: Since the leading term is −x3-x^3 (negative coefficient), the graph goes from the second quadrant to the fourth quadrant.

Explanation:

Expanding the roots helps identify the leading term and the constant dd. Plotting the intercepts first provides the skeleton for the cubic curve.

Problem 5:

Given the logarithmic function g(x)=ln⁡(x+2)g(x) = \ln(x + 2), determine the vertical asymptote and the xx-intercept.

Natural log graph shifted left by 2 units, showing asymptote at x = -2 and x-intercept at -1.

Solution:

  1. The vertical asymptote occurs where the argument of the log is zero: x+2=0  ⟹  x=−2x + 2 = 0 \implies x = -2.
  2. To find the xx-intercept, set g(x)=0g(x) = 0: ln⁡(x+2)=0  ⟹  e0=x+2  ⟹  1=x+2  ⟹  x=−1\ln(x + 2) = 0 \implies e^0 = x + 2 \implies 1 = x + 2 \implies x = -1.

Explanation:

The vertical asymptote marks the domain boundary. The x-intercept is found by solving the logarithmic equation for when the result is zero.