Given: $f(x) = -ax^2 + bx + 6$ 4.1 The gradient of the tangent to the graph of $f$ at the point $(-1, \frac{7}{2})$ is 3 - NSC Mathematics - Question 4 - 2017 - Paper 1

To calculate the $x$-intercepts, set $f(x) = 0$:

$-\frac{1}{2}x^2 + 2x + 6 = 0$

Multiplying through by -2 to eliminate the fraction:

$x^2 - 4x - 12 = 0$

Factoring gives:

$(x - 6)(x + 2) = 0$

Thus, the $x$-intercepts are:

$x = 6 \text{ and } x = -2.$

The turning point can be found using the formula:

$x = -\frac{b}{2a}$

Substituting the values:

$x = -\frac{2}{2(\frac{1}{2})} = -2.$

Now, find $f(-2)$:

$f(-2) = -\frac{1}{2}(-2)^2 + 2(-2) + 6 = -2 - 4 + 6 = 0.$

Thus, the coordinates of the turning point are $( -2, 0)$.

The graph of $f$ is a downward-opening parabola. The $x$-intercepts are $(6, 0)$ and $(-2, 0)$. The $y$-intercept is $f(0) = 6$, giving the point $(0, 6)$. The turning point is located at $(-2, 0)$. Make sure to clearly indicate these points on the graph.

From the sketch of the graph, observe where the parabola is above the line $y = 6$. The $x$-values for which $f(x) > 6$ can be determined by finding the regions where the graph exceeds the horizontal line $y = 6$. This will occur between the points where the parabola intersects the line $y = 6$; thus the answer will be an interval on the $x$-axis.

The line $g(x) = -x - 1$ has a $y$-intercept at $(0, -1)$ and $x$-intercept at $(-1, 0)$. Plot these points and draw the line extending through them, ensuring to mark the intercepts clearly on the graph.

To determine where the product $f(x)g(x) \leq 0$, examine the intervals where $f(x)$ and $g(x)$ intersect and their signs. Set up the necessary inequalities based on the graph to find the respective intervals.