The graph of h(x) passes through the point (0, -2) - Leaving Cert Mathematics - Question c - 2021

To find the maximum positive value of the slope of a tangent to h(x), we need to find the critical points of h'(x):

$h'(x) = -2x^2 + 4x + 6$

Setting the derivative to zero gives:

$-2x^2 + 4x + 6 = 0$

We can solve this using the quadratic formula, where:

• a = -2
• b = 4
• c = 6

The solutions are:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4(-2)(6)}}{2(-2)}$

$= \frac{-4 \pm \sqrt{16 + 48}}{-4} = \frac{-4 \pm 8}{-4}$

Calculating gives:

$x_1 = 1, \quad x_2 = -3$

To find the maximum slope, we evaluate h'(x) at these points:

• At x = 1:

$h'(1) = -2(1)^2 + 4(1) + 6 = -2 + 4 + 6 = 8$

• At x = -3:

$h'(-3) = -2(-3)^2 + 4(-3) + 6 = -18 - 12 + 6 = -24$

The maximum positive value of the slope of a tangent to h(x) occurs at x = 1, giving us a maximum slope of 8.