Given: $$ f(x)=-a x^{2}+b x+6 $$ 4.1 The gradient of the tangent to the graph of $$ f $$ at the point $$ \left(-1 ; \frac{7}{2}\right) $$ is 3 . Show that $$ a=\frac{1}{2} $$ and $$ b=2 $$. 2 Calculate the $$ x $$-intercepts of $$ f $$. Calculate the coordinates of the turning point of $$ f $$. Sketch the graph of $$ f $$. Clearly indicate ALL intercepts with the axes and the turning point. Use the graph to determine the values of $$ x $$ for which $$ f(x)6 $$. Sketch the graph of $$ g(x)=-x-1 $$ on the same set of axes as $$ f $$. Clearly indicate ALL intercepts with the axes. Write down the values of $$ x $$ for which $$ f(x) . g(x) \leq 0 $$. (3) (23)

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Given: $$ f(x)=-a x^{2}+b x+6 $$ 4.1 The gradient of the tangent to the graph of $$ f $$ at the point $$ \left(-1 ; \frac{7}{2}\right) $$ is 3 . Show that $$ a=\frac{1}{2} $$ and $$ b=2 $$. 2 Calculate the $$ x $$-intercepts of $$ f $$. Calculate the coordinates of the turning point of $$ f $$. Sketch the graph of $$ f $$. Clearly indicate ALL intercepts with the axes and the turning point. Use the graph to determine the values of $$ x $$ for which $$ f(x)>6 $$. Sketch the graph of $$ g(x)=-x-1 $$ on the same set of axes as $$ f $$. Clearly indicate ALL intercepts with the axes. Write down the values of $$ x $$ for which $$ f(x) . g(x) \leq 0 $$. (3) (23)

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Let's solve the problem step by step. ### Step 1: Show that $$ a = \frac{1}{2} $$ and $$ b = 2 $$ We are given the function: $$ f(x) = -a x^2 + b x + 6 $$ The gradient of the tangent to the graph of $$ f $$ is given by the derivative $$ f'(x) $$: $$ f'(x) = -2a x + b $$ We need to find the gradient at the point $$ \left(-1, \frac{7}{2}\right) $$. First, we substitute $$ x = -1 $$ into the derivative: $$ f'(-1) = -2a(-1) + b = 2a + b $$ We know that the gradient at this point is 3, so we set up the equation: $$ 2a + b = 3 \quad (1) $$ Next, we need to ensure that the function $$ f(-1) = \frac{7}{2} $$: $$ f(-1) = -a(-1)^2 + b(-1) + 6 = -a - b + 6 $$ Setting this equal to $$ \frac{7}{2} $$: $$ -a - b + 6 = \frac{7}{2} $$ Rearranging gives: $$ -a - b = \frac{7}{2} - 6 = \frac{7}{2} - \frac{12}{2} = -\frac{5}{2} $$ Thus, we have: $$ -a - b = -\frac{5}{2} \quad (2) $$ Now we have a system of equations: 1. $$ 2a + b = 3 $$ 2. $$ -a - b = -\frac{5}{2} $$ Let's solve this system of equations. ### Step 2: Solve the system of equations We can add equations (1) and (2): $$ (2a + b) + (-a - b) = 3 - \frac{5}{2} $$ This simplifies to: $$ a = 3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2} $$ Now substitute $$ a = \frac{1}{2} $$ back into equation (1): $$ 2\left(\frac{1}{2}\right) + b = 3 $$ This simplifies to: $$ 1 + b = 3 \implies b = 2 $$ Thus, we have shown that: $$ a = \frac{1}{2}, \quad b = 2 $$ ### Step 3: Calculate the $$ x $$-intercepts of $$ f $$ The $$ x $$-intercepts occur when $$ f(x) = 0 $$: $$ f(x) = -\frac{1}{2}x^2 + 2x + 6 = 0 $$ Multiplying through by -2 to eliminate the fraction: $$ x^2 - 4x - 12 = 0 $$ Now we can use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} $$ This gives us: $$ x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{-4}{2} = -2 $$ Thus, the $$ x $$-intercepts are $$ (6, 0) $$ and $$ (-2, 0) $$. ### Step 4: Calculate the coordinates of the turning point of $$ f $$ The turning point (vertex) of a quadratic function $$ ax^2 + bx + c $$ occurs at: $$ x = -\frac{b}{2a} $$ Substituting $$ a = -\frac{1}{2} $$ and $$ b = 2 $$: $$ x = -\frac{2}{2 \cdot -\frac{1}{2}} = -\frac{2}{-1} = 2 $$ Now, we find $$ f(2) $$: $$ f(2) = -\frac{1}{2}(2^2) + 2(2) + 6 = -\frac{1}{2}(4) + 4 + 6 = -2 + 4 + 6 = 8 $$ Thus, the turning point is $$ (2, 8) $$. ### Step 5: Sketch the graph of $$ f $$ The graph of $$ f $$ is a downward-opening parabola with: - $$ x $$-intercepts at $$ (6, 0) $$ and $$ (-2, 0) $$ - Turning point at $$ (2, 8) $$ ### Step 6: Determine the values of $$ x $$ for which $$ f(x) > 6 $$ We need to solve: $$ -\frac{1}{2}x^2 + 2x + 6 > 6 $$ This simplifies to: $$ -\frac{1}{2}x^2 + 2x > 0 $$ Factoring out $$ -\frac{1}{2} $$: $$ -\frac{1}{2}(x^2 - 4x) > 0 \implies x(x - 4) < 0 $$ The critical points are $$ x = 0 $$ and $$ x = 4 $$. The intervals to test are $$ (-\infty, 0) $$, $$ (0, 4) $$, and $$ (4, \infty) $$: - For $$ x < 0 $$: $$ f(x) > 6 $$ (true) - For $$ 0 < x < 4 $$: $$ f(x) > 6 $$ (true) - For $$ x > 4 $$: $$ f(x) < 6 $$ (false) Thus, $$ f(x) > 6 $$ for $$ 0 < x < 4 $$. ### Step 7: Sketch the graph of $$ g(x) = -x - 1 $$ The $$ x $$-intercept of $$ g(x) $$ occurs when $$ g(x) = 0 $$: $$ -x - 1 = 0 \implies x = -1 $$ The $$ y $$-intercept occurs when $$ x = 0 $$: \[ g(0 **Part 1:** - $$ a = \frac{1}{2} $$ - $$ b = 2 $$ **Part 2:** - $$ x $$-intercepts: $$ (6, 0) $$ and $$ (-2, 0) $$ **Part 3:** - Turning point: $$ (2, 8) $$ **Part 4:** - Graph of $$ f(x) $$ is a downward-opening parabola with $$ x $$-intercepts at $$ (6, 0) $$ and $$ (-2, 0) $$, and a turning point at $$ (2, 8) $$. **Part 5:** - $$ f(x) > 6 $$ when $$ 0 < x < 4 $$ **Part 6:** - Graph of $$ g(x) = -x - 1 $$ with $$ x $$-intercept at $$ (-1, 0) $$ and $$ y $$-intercept at $$ (0, -1) $$. **Part 7:** - $$ f(x) \cdot g(x) \leq 0 $$ when $$ x \leq -1 $$ or $$ x \geq 0 $$.

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