Practice Questions: Sketching Cubic Graphs

$$\begin{align*} \text{Step 1: Y-intercept (let } x = 0\text{):} \\ f(0) &= 5 \\ \text{Y-intercept: } &(0; 5) \\ \\ \text{Step 2: X-intercepts (let } f(x) = 0\text{):} \\ &\text{Using the factor theorem, try small values:} \\ f(1) &= 1 - 3 - 9 + 5 = -6 \neq 0 \\ f(-1) &= -1 - 3 + 9 + 5 = 10 \neq 0 \\ f(5) &= 125 - 75 - 45 + 5 = 10 \neq 0 \\ &\text{(X-intercepts not easily factorable)} \\ \\ \text{Step 3: Stationary points} \\ f'(x) &= 3x^2 - 6x - 9 \\ &= 3(x^2 - 2x - 3) \\ &= 3(x - 3)(x + 1) \\ \\ f'(x) = 0 \text{ when } &x = 3 \text{ or } x = -1 \\ \\ f(3) &= 27 - 27 - 27 + 5 = -22 \\ f(-1) &= -1 - 3 + 9 + 5 = 10 \\ \\ \text{Step 4: Classify using } f''(x): \\ f''(x) &= 6x - 6 \\ f''(3) &= 12 > 0 \quad \text{(minimum)} \\ f''(-1) &= -12 < 0 \quad \text{(maximum)} \\ \\ \text{Step 5: Point of inflection} \\ f''(x) &= 0 \\ 6x - 6 &= 0 \\ x &= 1 \\ f(1) &= 1 - 3 - 9 + 5 = -6 \\ \text{Inflection point: } &(1; -6) \end{align*}$$

Summary for sketch:

• Y-intercept: $(0; 5)$
• Local maximum: $(-1; 10)$
• Point of inflection: $(1; -6)$
• Local minimum: $(3; -22)$
• As $x \to -\infty$, $f(x) \to -\infty$
• As $x \to +\infty$, $f(x) \to +\infty$

$$\begin{align*} \text{Step 1: Y-intercept} \\ g(0) &= 2 \\ \text{Point: } &(0; 2) \\ \\ \text{Step 2: Stationary points} \\ g'(x) &= -3x^2 + 3 \\ &= -3(x^2 - 1) \\ &= -3(x - 1)(x + 1) \\ \\ g'(x) = 0 \text{ when } &x = 1 \text{ or } x = -1 \\ \\ g(1) &= -1 + 3 + 2 = 4 \\ g(-1) &= 1 - 3 + 2 = 0 \\ \\ \text{Step 3: Classify} \\ g''(x) &= -6x \\ g''(1) &= -6 < 0 \quad \text{(maximum)} \\ g''(-1) &= 6 > 0 \quad \text{(minimum)} \\ \\ \text{Step 4: Check behaviour} \\ &\text{Leading coefficient is negative } (-x^3) \\ &\text{So: } x \to -\infty, g(x) \to +\infty \\ &\phantom{\text{So: }} x \to +\infty, g(x) \to -\infty \end{align*}$$

Summary for sketch:

• Y-intercept: $(0; 2)$
• X-intercept: $(-1; 0)$
• Local minimum: $(-1; 0)$
• Local maximum: $(1; 4)$
• Graph starts high (top left) and ends low (bottom right)

$$\begin{align*} \text{Step 1: Y-intercept} \\ h(0) &= 0 \\ \text{Point: } &(0; 0) \\ \\ \text{Step 2: X-intercepts} \\ x^3 - 6x^2 + 9x &= 0 \\ x(x^2 - 6x + 9) &= 0 \\ x(x - 3)^2 &= 0 \\ x = 0 \text{ or } &x = 3 \text{ (repeated)} \\ \\ \text{Step 3: Stationary points} \\ h'(x) &= 3x^2 - 12x + 9 \\ &= 3(x^2 - 4x + 3) \\ &= 3(x - 1)(x - 3) \\ \\ h'(x) = 0 \text{ when } &x = 1 \text{ or } x = 3 \\ \\ h(1) &= 1 - 6 + 9 = 4 \\ h(3) &= 27 - 54 + 27 = 0 \\ \\ \text{Step 4: Classify} \\ h''(x) &= 6x - 12 \\ h''(1) &= -6 < 0 \quad \text{(maximum)} \\ h''(3) &= 6 > 0 \quad \text{(minimum)} \\ \\ \text{Step 5: Point of inflection} \\ h''(x) &= 0 \\ 6x - 12 &= 0 \\ x &= 2 \\ h(2) &= 8 - 24 + 18 = 2 \\ \text{Inflection point: } &(2; 2) \end{align*}$$

Summary for sketch:

• Intercepts: $(0; 0)$ and $(3; 0)$
• Local maximum: $(1; 4)$
• Point of inflection: $(2; 2)$
• Local minimum (also x-intercept): $(3; 0)$
• Note: At $x = 3$, the graph touches but doesn't cross the x-axis (repeated root)

$$\begin{align*} \text{Step 1: Find } f'(x) \\ f'(x) &= 6x^2 - 18x + 12 \\ &= 6(x^2 - 3x + 2) \\ &= 6(x - 1)(x - 2) \\ \\ \text{Step 2: Solve } f'(x) = 0 \\ x &= 1 \text{ or } x = 2 \\ \\ \text{Step 3: Find y-coordinates} \\ f(1) &= 2(1) - 9(1) + 12(1) - 4 = 1 \\ f(2) &= 2(8) - 9(4) + 12(2) - 4 = 0 \\ \\ \text{Step 4: Classify using } f''(x) \\ f''(x) &= 12x - 18 \\ f''(1) &= 12 - 18 = -6 < 0 \quad \text{(maximum)} \\ f''(2) &= 24 - 18 = 6 > 0 \quad \text{(minimum)} \end{align*}$$

Answers:

• Local maximum at $(1; 1)$
• Local minimum at $(2; 0)$

$$\begin{align*} \text{Step 1: Intercepts} \\ \text{Y-intercept: } p(0) &= 0 \\ \\ \text{X-intercepts:} \\ -2x^3 + 6x^2 &= 0 \\ -2x^2(x - 3) &= 0 \\ x = 0 \text{ or } &x = 3 \\ \\ \text{Step 2: Stationary points} \\ p'(x) &= -6x^2 + 12x \\ &= -6x(x - 2) \\ \\ p'(x) = 0 \text{ when } &x = 0 \text{ or } x = 2 \\ \\ p(0) &= 0 \\ p(2) &= -2(8) + 6(4) = 8 \\ \\ \text{Step 3: Classify} \\ p''(x) &= -12x + 12 \\ p''(0) &= 12 > 0 \quad \text{(minimum)} \\ p''(2) &= -12 < 0 \quad \text{(maximum)} \\ \\ \text{Step 4: Point of inflection} \\ p''(x) &= 0 \\ -12x + 12 &= 0 \\ x &= 1 \\ p(1) &= -2 + 6 = 4 \end{align*}$$

Summary for sketch:

• Intercepts: $(0; 0)$ and $(3; 0)$
• Local minimum: $(0; 0)$
• Point of inflection: $(1; 4)$
• Local maximum: $(2; 8)$
• Leading coefficient is negative, so graph falls to the right

$$\begin{align*} \text{Find } f''(x): \\ f'(x) &= 3x^2 - 12 \\ f''(x) &= 6x \\ \\ \text{Set } f''(x) = 0: \\ 6x &= 0 \\ x &= 0 \\ \\ \text{Find y-coordinate:} \\ f(0) &= 0 - 0 + 1 = 1 \end{align*}$$

Answer: Point of inflection at $(0; 1)$

Verification: $f''(x) = 6x$ changes sign at $x = 0$ (from negative to positive), confirming this is a point of inflection.

$$\begin{align*} \text{Step 1: Expand} \\ k(x) &= (x + 1)(x - 2)^2 \\ &= (x + 1)(x^2 - 4x + 4) \\ &= x^3 - 4x^2 + 4x + x^2 - 4x + 4 \\ &= x^3 - 3x^2 + 4 \\ \\ \text{Step 2: Intercepts} \\ \text{Y-intercept: } k(0) &= 4 \\ \\ \text{X-intercepts:} \\ (x + 1)(x - 2)^2 &= 0 \\ x = -1 \text{ or } &x = 2 \text{ (repeated)} \\ \\ \text{Step 3: Stationary points} \\ k'(x) &= 3x^2 - 6x \\ &= 3x(x - 2) \\ \\ k'(x) = 0 \text{ when } &x = 0 \text{ or } x = 2 \\ \\ k(0) &= 4 \\ k(2) &= 0 \\ \\ \text{Step 4: Classify} \\ k''(x) &= 6x - 6 \\ k''(0) &= -6 < 0 \quad \text{(maximum)} \\ k''(2) &= 6 > 0 \quad \text{(minimum)} \end{align*}$$

Summary for sketch:

• Y-intercept: $(0; 4)$
• X-intercepts: $(-1; 0)$ and $(2; 0)$ (repeated root)
• Local maximum: $(0; 4)$
• Local minimum (also x-intercept): $(2; 0)$
• At $x = 2$, graph touches but doesn't cross x-axis

$$\begin{align*} \text{Find } f''(x): \\ f'(x) &= 3x^2 - 12x + 9 \\ f''(x) &= 6x - 12 \\ \\ \text{For concave up, we need } f''(x) > 0: \\ 6x - 12 &> 0 \\ 6x &> 12 \\ x &> 2 \end{align*}$$

Answer: The graph is concave up when $x > 2$.

Note: The graph is concave down when $x < 2$, and changes concavity at $x = 2$ (the point of inflection).

$$\begin{align*} \text{For a point of inflection:} \\ g''(x) &= 0 \\ \\ \text{Find } g''(x): \\ g'(x) &= 3x^2 + 2ax \\ g''(x) &= 6x + 2a \\ \\ \text{At } x = -2: \\ g''(-2) &= 0 \\ 6(-2) + 2a &= 0 \\ -12 + 2a &= 0 \\ a &= 6 \\ \\ \text{The point } (-2; 5) \text{ lies on the curve:} \\ g(-2) &= 5 \\ (-2)^3 + 6(-2)^2 + b &= 5 \\ -8 + 24 + b &= 5 \\ 16 + b &= 5 \\ b &= -11 \end{align*}$$

Answers: $a = 6$ and $b = -11$

$$\begin{align*} \text{Since } (1; 0) \text{ and } (3; -4) \text{ are turning points:} \\ f'(1) &= 0 \text{ and } f'(3) = 0 \\ \\ f'(x) &= 3x^2 + 2px + q \\ \\ \text{Using } f'(1) = 0: \\ 3(1)^2 + 2p(1) + q &= 0 \\ 3 + 2p + q &= 0 \quad \text{...(1)} \\ \\ \text{Using } f'(3) = 0: \\ 3(3)^2 + 2p(3) + q &= 0 \\ 27 + 6p + q &= 0 \quad \text{...(2)} \\ \\ \text{Subtract (1) from (2):} \\ 24 + 4p &= 0 \\ p &= -6 \\ \\ \text{Substitute into (1):} \\ 3 + 2(-6) + q &= 0 \\ 3 - 12 + q &= 0 \\ q &= 9 \\ \\ \text{Using point } (1; 0): \\ f(1) &= 0 \\ 1 + p + q + r &= 0 \\ 1 - 6 + 9 + r &= 0 \\ r &= -4 \\ \\ \text{Check with } (3; -4): \\ f(3) &= 27 + (-6)(9) + 9(3) + (-4) \\ &= 27 - 54 + 27 - 4 = -4 \checkmark \end{align*}$$

Answers: $p = -6$, $q = 9$, $r = -4$