The graph of $f(x)=(x+4)(x-6)$ is drawn below - NSC Mathematics - Question 8 - 2021 - Paper 1

To find the y-intercept, we set $x = 0$ in the function:

$f(0) = (0+4)(0-6) = 4(-6) = -24$

Thus, the y-intercept is $G(0, -24)$.

To find the turning point of the parabola, we can use the vertex formula for a quadratic in the form $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex. The midpoint between the x-intercepts gives:

$h = \frac{-4 + 6}{2} = 1$

Substituting $h$ back into the function to find $k$:

$f(1) = (1 + 4)(1 - 6) = 5(-5) = -25$

Thus, the turning point $C(1, -25)$.

The angle of inclination $\theta$ can be found using the slope of the line $AE$. The slope is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

The coordinates of point $E$ are $(0, -24)$ and point $A$ can be determined further based on the context provided on the graph. The angle can then be calculated using:

$\tan(\theta) = m$

Using the arctangent function, we can find $\theta$.

Point $T$ is a point on the parabola between points $B$ and $G$. We can find $T$ by selecting a value of $x$ between $-4$ and $0$. For instance, choosing $x=-2$:

$f(-2) = (-2 + 4)(-2 - 6) = 2(-8) = -16$

Therefore, point $T(-2, -16)$ is valid.