Find $$\int (4x^3 - 6x + 10) \, dx.$$ Part of the graph of a cubic function $f(x)$ is shown below (graph not to scale) - Leaving Cert Mathematics - Question 4 - 2019

To find $f(x)$, we need to integrate $f'(x)$:

$\int (6x^2 - 54x + 109) \, dx = 2x^3 - 27x^2 + 109x + C$

Next, we need to find the constant $C$. We know that as $f(2) = 0$:

For $x = 2$: $f(2) = 2(2)^3 - 27(2)^2 + 109(2) + C = 0$

Calculating this gives: $2(8) - 27(4) + 218 + C = 0$ $16 - 108 + 218 + C = 0$ $126 + C = 0$ Thus, $C = -126$

Finally, we find: $f(x) = 2x^3 - 27x^2 + 109x - 126$

To find the points where the graph intersects the x-axis, we must solve:

$f(x) = 0$

From the earlier expression, we have: $2x^3 - 27x^2 + 109x - 126 = 0$

Using synthetic division or the factor theorem, we check for rational roots, finding that $x = 2$ is a root. Thus, $(x - 2)$ is a factor:

Dividing the cubic polynomial by $(x - 2)$:

We find: $2x^3 - 27x^2 + 109x - 126 = (x - 2)(2x^2 - 23x + 63)$

Next, we resolve the quadratic equation: $2x^2 - 23x + 63 = 0$ By using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Which gives: $x = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 2 \cdot 63}}{2 \cdot 2}$ $= \frac{23 \pm \sqrt{529 - 504}}{4}$ $= \frac{23 \pm \sqrt{25}}{4}$ $= \frac{23 \pm 5}{4}$

Thus: $x = \frac{28}{4} = 7\quad \text{or} \quad x = \frac{18}{4} = 4.5$

So the coordinates for points B and C are: $B = (4.5, 0) \quad C = (7, 0)$