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Algebra - Representation and shape of more complex functions-extended

Grade 10IB

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

🔑Concepts

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The cubic function y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d can have up to two turning points (a local maximum and a local minimum) and cross the x-axis at most three times. The 'S' shape depends on the sign of aa; if a>0a > 0, the graph starts low and ends high, whereas if a<0a < 0, it starts high and ends low.

Graph of a cubic function showing local maximum and minimum.
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Reciprocal functions of the form y=kx−h+vy = \frac{k}{x-h} + v are hyperbolas with two branches. They feature a vertical asymptote at x=hx = h (where the denominator is zero) and a horizontal asymptote at y=vy = v (the value yy approaches as xx becomes very large).

Graph of y=1/x showing horizontal and vertical asymptotes at the axes.
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Exponential functions y=a⋅bx+vy = a \cdot b^x + v (where b>0,b≠1b > 0, b \neq 1) represent rapid growth (b>1b > 1) or decay (0<b<10 < b < 1). These functions have a horizontal asymptote at y=vy = v and never touch or cross it. The y-intercept occurs at (0,a+v)(0, a + v).

Exponential growth curve showing y-intercept at (0,1).
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Absolute value functions y=∣x−h∣+vy = |x - h| + v create a distinct 'V' shape. The vertex of the graph is located at (h,v)(h, v). The graph is symmetric about the vertical line x=hx = h.

V-shaped graph of the absolute value function.

📐Formulae

y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d

y=kx−h+vy = \frac{k}{x - h} + v

y=a⋅bx−h+vy = a \cdot b^{x-h} + v

x=h (Vertical Asymptote for y=kx−h)x = h \text{ (Vertical Asymptote for } y = \frac{k}{x-h})

y=v (Horizontal Asymptote for y=kx−h or y=bx+v)y = v \text{ (Horizontal Asymptote for } y = \frac{k}{x-h} \text{ or } y = b^x + v)

💡Examples

Problem 1:

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

Solution:

To find the yy-intercept, set x=0x = 0: y=(0)3−4(0)=0y = (0)^3 - 4(0) = 0 So, the yy-intercept is (0,0)(0, 0).

To find the xx-intercepts, set y=0y = 0: 0=x3−4x0 = x^3 - 4x 0=x(x2−4)0 = x(x^2 - 4) 0=x(x−2)(x+2)0 = x(x - 2)(x + 2) Solving for xx, we get x=0x = 0, x=2x = 2, and x=−2x = -2.

Explanation:

We use factoring to solve the cubic equation. Setting y=0y=0 allows us to find where the graph crosses the horizontal axis.

Problem 2:

Identify the equations of the asymptotes for the reciprocal function y=3x+2−5y = \frac{3}{x + 2} - 5.

Solution:

The vertical asymptote occurs where the denominator is zero: x+2=0  ⟹  x=−2x + 2 = 0 \implies x = -2 The horizontal asymptote is determined by the constant term added to the fraction: y=−5y = -5

Explanation:

In the form y=kx−h+vy = \frac{k}{x-h} + v, the vertical asymptote is x=hx=h and the horizontal asymptote is y=vy=v.

Problem 3:

A population of bacteria doubles every hour. If the initial population is 50, write a function P(t)P(t) to represent the population after tt hours and find the population after 3 hours.

Solution:

The general form for exponential growth is P(t)=P0⋅btP(t) = P_0 \cdot b^t. Initial population P0=50P_0 = 50, growth factor b=2b = 2. P(t)=50⋅2tP(t) = 50 \cdot 2^t For t=3t = 3: P(3)=50⋅23P(3) = 50 \cdot 2^3 P(3)=50⋅8=400P(3) = 50 \cdot 8 = 400

Explanation:

We model the growth using an exponential function where the base represents the doubling effect and the coefficient represents the starting value.

Problem 4:

Sketch the graph of y=(x−2)3+1y = (x - 2)^3 + 1 and label the point of inflection.

Cubic graph with point of inflection at (2,1).

Solution:

  1. Identify the parent function: y=x3y = x^3.
  2. Identify transformations: The graph is shifted 2 units to the right (h=2h = 2) and 1 unit up (v=1v = 1).
  3. The point of inflection for y=x3y = x^3 is (0,0)(0,0). For y=(x−2)3+1y = (x - 2)^3 + 1, the point of inflection is (2,1)(2, 1).
  4. Calculate yy-intercept: y=(0−2)3+1=−8+1=−7y = (0 - 2)^3 + 1 = -8 + 1 = -7.

Explanation:

Translating a cubic function involves moving the point where the curvature changes (the point of inflection) according to the constants added to xx and the function value.

Problem 5:

Determine the horizontal and vertical asymptotes for y=1x−3+2y = \frac{1}{x-3} + 2 and sketch the graph.

Reciprocal function graph with asymptotes at x=3 and y=2.

Solution:

  1. Vertical Asymptote: Set the denominator to zero: x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3.
  2. Horizontal Asymptote: As x→∞x \to \infty, 1x−3→0\frac{1}{x-3} \to 0, so y→2y \to 2. The horizontal asymptote is y=2y = 2.
  3. Intercepts: yy-intercept: y=10−3+2=123y = \frac{1}{0-3} + 2 = 1\frac{2}{3}. xx-intercept: 0=1x−3+2  ⟹  −2=1x−3  ⟹  x−3=−0.5  ⟹  x=2.50 = \frac{1}{x-3} + 2 \implies -2 = \frac{1}{x-3} \implies x-3 = -0.5 \implies x = 2.5.

Explanation:

Asymptotes act as boundaries that the curve approaches but never reaches. The vertical shift vv moves the horizontal asymptote, and the horizontal shift hh moves the vertical asymptote.