Given: $f(x) = 2x^3 - 5x^2 + 4x$ 8.1 Calculate the coordinates of the turning points of the graph of $f$ - NSC Mathematics - Question 8 - 2017 - Paper 1

Factoring out $x$ gives:

$x(2x^2 - 5x + 4) = 0$

This has a root at $x = 0$.

Now, for the quadratic $2x^2 - 5x + 4 = 0$, applying the discriminant method:

$D = b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 4 = 25 - 32 = -7$

Since the discriminant is negative, the quadratic has no real roots. Therefore, the only real root of the original equation is at $x = 0$.

To sketch the graph:

1. Intercepts with the axes:

   • x-intercept: $x = 0$ (as found earlier).
   • y-intercept: Calculate $f(0)$, which yields $f(0) = 0$. Thus, the intercept is at $(0, 0)$.

2. Turning Points:

   • $(1, 1)$ and $\left(\frac{2}{3}, \frac{28}{27}\right)$.

3. Graph Characteristics:

   • The function is cubic, hence it extends to infinity as $x \to \pm \infty$.
   • The graph has three critical points and will change direction around them. It will be increasing towards and decreasing after the turning points.

To determine concavity, we first compute the second derivative:

$f''(x) = 12x - 10$

Setting this to zero to find points of inflection:

$12x - 10 = 0 \Rightarrow x = \frac{5}{6}$

Next, we check intervals around this point:

• For $x < \frac{5}{6}$, choose $x = 0$:
  $f''(0) = -10 < 0\quad \Rightarrow\text{ concave down.}$
• For $x > \frac{5}{6}$, choose $x = 1$:
  $f''(1) = 2 > 0\quad \Rightarrow\text{ concave up.}$

Therefore, the graph of $f$ is concave up for $x > \frac{5}{6}$.