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)

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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)

Answer 1

To solve the given problems, let's analyze each part step by step.### 9.1.1 Write down the coordinates of point E.The y-intercept is where

. Since both \( g'(x) \) and \( h(x) \) have a common y-intercept at point E, we can find it by evaluating both functions at

.For \( h(x) = 2x - 4 \):

Thus, the coordinates of point E are \((0, -4)\).### 9.1.2 Determine the equation of the graph of

in the form

.Given that the x-intercepts of \( g'(x) \) are \( C(-2, 0) \) and \( D(6, 0) \), we can write:

Since the y-intercept is \((0, -4)\), substitute

into the equation:

Thus, the equation becomes:

### 9.1.3 Write down the x-coordinates of the turning point of

.The turning points of

occur where \( g'(x) = 0 \), which are the x-intercepts of

:

and

.### 9.1.4 Write down the x-coordinate of the point of inflection of the graph of

.The point of inflection occurs where the second derivative \( g''(x) = 0 \). The turning point of

is also the point of inflection for

.To find the turning point of \( g'(x) \), use the vertex formula for a quadratic function:

Thus, the x-coordinate of the point of inflection of

is

.### 9.1.5 Explain why

has a local maximum at

.Since \( g'(x) \) changes from positive to negative at

, this indicates a local maximum for

at

.### 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.To determine if a tangent with a negative gradient is possible, find the derivative \( h'(x) \):

Since

is always positive (as

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.

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