The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below - NSC Mathematics - Question 9 - 2021 - Paper 1

To find the turning points, we first compute the first derivative:

$f'(x) = 6x^2 + 6x - 12$

Next, we set the first derivative equal to zero to find critical points:

$6x^2 + 6x - 12 = 0$

Dividing by 6 gives:

$x^2 + x - 2 = 0$

Factoring the quadratic, we get:

$(x - 1)(x + 2) = 0$

Thus, the solutions are:

$x = -2 \text{ and } x = 1$

Now, we will find the corresponding $y$ values by substituting back into the original function:

For $x = -2$:

$f(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) = 20$

For $x = 1$:

$f(1) = 2(1)^3 + 3(1)^2 - 12(1) = -7$

Thus, the coordinates are:

$A(-2; 20) ext{ and } B(1; -7)$