Calculate the coordinates of the turning point of the graph of f. Calculate the coordinates of P, the point of contact between f and g. Hence or otherwise, determine the equation of g. Determine the values of d for which the line k(x)=-x+d will not intersect the graph of f.

Here are the solutions to Question 5:

5.1 Calculate the coordinates of the turning point of the graph of $f$. The function is $f(x) = -2x^2 - 5x + 3$.

Step 1: Find the x-coordinate of the turning point using the formula $x = -\frac{b}{2a}$. $x = -\frac{-5}{2(-2)} = -\frac{-5}{-4} = -\frac{5}{4}$

Step 2: Substitute the x-coordinate back into $f(x)$ to find the y-coordinate. $f\left(-\frac{5}{4}\right) = -2\left(-\frac{5}{4}\right)^2 - 5\left(-\frac{5}{4}\right) + 3$ $f\left(-\frac{5}{4}\right) = -2\left(\frac{25}{16}\right) + \frac{25}{4} + 3$ $f\left(-\frac{5}{4}\right) = -\frac{25}{8} + \frac{50}{8} + \frac{24}{8} = \frac{49}{8}$ The coordinates of the turning point are $\boxed{\left(-\frac{5}{4}, \frac{49}{8}\right)}$.

5.2 Calculate the coordinates of P, the point of contact between $f$ and $g$. The angle of inclination of graph $g$ is $135^\circ$. The line $g$ is tangent to $f$ at P.

Step 1: Determine the gradient of $g(x)$ using the angle of inclination. $m = \tan(\theta) = \tan(135^\circ) = -1$

Step 2: Find the derivative of $f(x)$. $f'(x) = \frac{d}{dx}(-2x^2 - 5x + 3) = -4x - 5$

Step 3: At the point of tangency P, the gradient of $f(x)$ is equal to the gradient of $g(x)$. $-4x_P - 5 = -1$ $-4x_P = 4$ $x_P = -1$

Step 4: Substitute $x_P = -1$ into $f(x)$ to find the y-coordinate of P. $f(-1) = -2(-1)^2 - 5(-1) + 3$ $f(-1) = -2(1) + 5 + 3 = 6$ The coordinates of P are $\boxed{(-1, 6)}$.

5.3 Hence or otherwise, determine the equation of $g$. We know $g(x) = ax + q$. From 5.2, the gradient $a = -1$ and the point P is $(-1, 6)$.

Step 1: Substitute the gradient and the coordinates of P into the equation $g(x) = ax + q$. $6 = (-1)(-1) + q$ $6 = 1 + q$ $q = 5$

Step 2: Write the equation of $g(x)$. $g(x) = -x + 5$ The equation of $g$ is $\boxed{g(x) = -x + 5}$.

5.4 Determine the values of $d$ for which the line $k(x) = -x + d$ will not intersect the graph of $f$. For the line $k(x)$ not to intersect $f(x)$, the equation $f(x) = k(x)$ must have no real solutions.

Step 1: Set $f(x) = k(x)$ and rearrange into a quadratic equation. $-2x^2 - 5x + 3 = -x + d$ $-2x^2 - 4x + (3 - d) = 0$

Step 2: For no real solutions, the discriminant $\Delta = B^2 - 4AC$ must be less than zero. Here, $A = -2$, $B = -4$, $C = (3 - d)$. $(-4)^2 - 4(-2)(3 - d) < 0$ $16 + 8(3 - d) < 0$ $16 + 24 - 8d < 0$ $40 - 8d < 0$

Step 3: Solve the inequality for $d$. $40 < 8d$ $d > \frac{40}{8}$ $d > 5$ The values of $d$ for which the line $k(x)$ will not intersect $f(x)$ are $\boxed{d > 5}$.