Given: $f(x) = x^3 + 4x^2 - 7x - 10$ 8.1 Write down the y-intercept of $f$ - NSC Mathematics - Question 8 - 2023 - Paper 1

To verify that $2$ is a root, we substitute $x = 2$ into the function $f(x)$:

$f(2) = 2^3 + 4(2)^2 - 7(2) - 10 = 8 + 16 - 14 - 10 = 0.$

Since $f(2) = 0$, we confirm that $2$ is indeed a root.

Since $2$ is a root, $(x - 2)$ is a factor of $f(x)$.

Using synthetic division or polynomial long division:

$f(x) = (x - 2)(x^2 + 6x + 5).$

Factoring the quadratic gives:

$x^2 + 6x + 5 = (x + 5)(x + 1).$

Hence, the complete factorisation is:

$f(x) = (x - 2)(x + 5)(x + 1).$

To sketch the graph, plot the y-intercept at $(0, -10)$ and the turning points at $(0.7; 12.6)$ and $(-3.4; 20.8)$.

Draw the curve passing through these points while ensuring that it intercepts the x-axis at the roots: $(2, 0)$, $(-5, 0)$, and $(-1, 0)$.

The graph should show a general cubic shape, with the curve rising and falling at appropriate intervals.

From the graph, $f^{ ext{'}}(x) < 0$ for the intervals $x ext{ in } (- ext{∞}, -3.4)$ and $(0.7, + ext{∞})$.

The minimum gradient occurs at the turning points. Therefore, $f^{ ext{'}}(x)$ will achieve its minimum at the x-coordinates of the turning points: $x = -3.4$ and $x=0.7$.

For the expression $f^{ ext{'}}(g) imes f^{ ext{'}}(x) eq 0$, we need to identify the x-values where $f^{ ext{'}}(x)$ changes sign, which occurs on intervals outside of the turning points, specifically having x-values in the intervals (- ext{∞}, -3.4) igcup (-3.4; 0.7).