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 show that $2$ is a root, we evaluate $f(2)$:

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

Since $f(2) = 0$, $2$ is indeed a root of the equation.

Given that $2$ is a root, we can factor $f(x)$ as follows:

$f(x) = (x - 2)(Ax^2 + Bx + C).$

Using polynomial long division or synthetic division, we find:

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

Factoring $x^2 + 6x + 5$ gives:

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

Thus, the complete factorization is: $f(x) = (x - 2)(x + 5)(x + 1).$

To sketch the graph of $f(x)$, we can plot the following points:

• Intercepts: (0, -10), (2, 0), (-5, 0), (-1, 0)
• Turning points: approx. (0.7, 12.6) and (-3.4, 20.8)

The sketch should show the function starting from the negative side, approaching the y-intercept, and having turning points where the curve changes direction.

To find where $f'(x) < 0$, we first calculate $f'(x)$:

$f'(x) = 3x^2 + 8x - 7.$

We need to solve for $f'(x) = 0$ to find the critical points. Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} o x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-7)}}{2(3)} \to x = \frac{-8 \pm \sqrt{64 + 84}}{6} \to x = \frac{-8 \pm \sqrt{148}}{6}.$

This gives two critical points. Evaluate intervals to determine where $f'(x)$ is negative.

To find where the gradient of the tangent is a minimum, we need to evaluate points where the second derivative $f''(x) = 0$ to find inflection points. Calculate:

$f''(x) = 6x + 8.$

Setting this to zero allows finding critical points, then testing intervals to ensure we find where the gradient is minimized.