QUESTION 9 9.1 The graphs of g'(x)=ax^2+bx+c and h(x)=2x-4 are sketched below. The graph of g'(x)=ax^2+bx+c is the derivative graph of a cubic function g. The graphs of h and g' have a common y-intercept at point E. C(-2;0) and D(6;0) are the x-intercepts of the graph of g' Point A is the x-intercept of h and point B is the turning point of g' Line AB is parallel to the y-axis. 9.1.1 Write down the coordinates of point E. (1) 9.1.2 Determine the equation of the graph of g'in the form y=ax^2+bx+c. (4) 9.1.3 Write down the x-coordinates of the turning point of g. (2) 9.1.4 Write down the x-coordinate of the point of inflection of the graph of g. (1) 9.1.5 Explain why g has a local maximum at x=-2. (2) 9.2 Given: h(x)=4x^3+5x Substantiate whether it is possible to draw a tangent to the graph of h that has a negative gradient. (2)

To solve the given problems, let's analyze each part step by step.<br /><br />### 9.1.1 Write down the coordinates of point E.<br />The y-intercept is where \( x = 0 \). Since both \( g'(x) \) and \( h(x) \) have a common y-intercept at point E, we can find it by evaluating both functions at \( x = 0 \).<br /><br />For \( h(x) = 2x - 4 \):<br />\[ h(0) = 2(0) - 4 = -4 \]<br /><br />Thus, the coordinates of point E are \((0, -4)\).<br /><br />### 9.1.2 Determine the equation of the graph of \( g' \) in the form \( y = ax^2 + bx + c \).<br />Given that the x-intercepts of \( g'(x) \) are \( C(-2, 0) \) and \( D(6, 0) \), we can write:<br />\[ g'(x) = a(x + 2)(x - 6) \]<br /><br />Since the y-intercept is \((0, -4)\), substitute \( x = 0 \) into the equation:<br />\[ g'(0) = a(0 + 2)(0 - 6) = -12a = -4 \]<br />\[ a = \frac{1}{3} \]<br /><br />Thus, the equation becomes:<br />\[ g'(x) = \frac{1}{3}(x + 2)(x - 6) \]<br />\[ g'(x) = \frac{1}{3}(x^2 - 4x - 12) \]<br />\[ g'(x) = \frac{1}{3}x^2 - \frac{4}{3}x - 4 \]<br /><br />### 9.1.3 Write down the x-coordinates of the turning point of \( g \).<br />The turning points of \( g \) occur where \( g'(x) = 0 \), which are the x-intercepts of \( g' \): \( x = -2 \) and \( x = 6 \).<br /><br />### 9.1.4 Write down the x-coordinate of the point of inflection of the graph of \( g \).<br />The point of inflection occurs where the second derivative \( g''(x) = 0 \). The turning point of \( g' \) is also the point of inflection for \( g \).<br /><br />To find the turning point of \( g'(x) \), use the vertex formula for a quadratic function:<br />\[ x = -\frac{b}{2a} = -\frac{-4/3}{2 \times 1/3} = 2 \]<br /><br />Thus, the x-coordinate of the point of inflection of \( g \) is \( x = 2 \).<br /><br />### 9.1.5 Explain why \( g \) has a local maximum at \( x = -2 \).<br />Since \( g'(x) \) changes from positive to negative at \( x = -2 \), this indicates a local maximum for \( g \) at \( x = -2 \).<br /><br />### 9.2 Substantiate whether it is possible to draw a tangent to the graph of \( h(x) = 4x^3 + 5x \) that has a negative gradient.<br />To determine if a tangent with a negative gradient is possible, find the derivative \( h'(x) \):<br />\[ h'(x) = 12x^2 + 5 \]<br /><br />Since \( 12x^2 + 5 \) is always positive (as \( 12x^2 \geq 0 \) and adding 5 keeps it positive), the gradient of any tangent line to \( h(x) \) is always positive. Therefore, it is not possible to draw a tangent with a negative gradient.