The graph of $f(x) = ax^3 + bx^2 + cx - 5$ is drawn below. E(-1 ; 0) and G(5 ; 0) are the x-intercepts of $f$. 9.1 Show that $a = 1$, $b = -3$ and $c = -9$. 9.2 Calculate the value of $x$ for which $f$ has a local minimum value. 9.3 Use the graph to determine the values of $x$ for which $f''(x) > 0$. 9.4 For which values of $c$ will the graph of $p(x) = f(x) + t$ have two distinct positive roots and one negative root?

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The graph of $f(x) = ax^3 + bx^2 + cx - 5$ is drawn below - NSC Mathematics - Question 9 - 2024 - Paper 1

Question 9

The graph of $f(x) = ax^3 + bx^2 + cx - 5$ is drawn below. E(-1 ; 0) and G(5 ; 0) are the x-intercepts of $f$. 9.1 Show that $a = 1$, $b = -3$ and $c = -9$. 9.... show full transcript

Worked Solution & Example Answer:The graph of $f(x) = ax^3 + bx^2 + cx - 5$ is drawn below - NSC Mathematics - Question 9 - 2024 - Paper 1

Step 1

Show that $a = 1$, $b = -3$ and $c = -9$.

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Answer

To find the coefficients, we substitute the known x-intercepts into the function. Since $E(-1; 0)$ and $G(5; 0)$ are x-intercepts, we have:

$f(-1) = a(-1)^3 + b(-1)^2 + c(-1) - 5 = 0$ which simplifies to:

$-a + b - c - 5 = 0$

For point $G(5; 0)$ :

$f(5) = a(5)^3 + b(5)^2 + c(5) - 5 = 0$ which simplifies to:

$125a + 25b + 5c - 5 = 0$

Now substituting a known value for $f(0) = -5$ at $x=0$ gives:

$f(0) = a(0)^3 + b(0)^2 + c(0) - 5 = -5$ From these equations, solving yields $a = 1$ , $b = -3$ , and $c = -9$ .

Step 2

Calculate the value of $x$ for which $f$ has a local minimum value.

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Answer

To find the local minimum, we set the first derivative $f'(x) = 3ax^2 + 2bx + c$ to zero. With $a = 1$ , $b = -3$ , and $c = -9$ , we have:

$f'(x) = 3x^2 - 6x - 9$ Setting $f'(x) = 0$ gives:

$3x^2 - 6x - 9 = 0$ Dividing the entire equation by 3:

$x^2 - 2x - 3 = 0$ By factoring, we find:

$(x - 3)(x + 1) = 0$ Thus, $x = 3$ or $x = -1$ . The second derivative test can confirm that $x = 3$ gives a local minimum.

Step 3

Use the graph to determine the values of $x$ for which $f''(x) > 0$.

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Answer

To find where $f''(x) > 0$ , we first compute the second derivative:

$f''(x) = 6ax - 6b$ Substituting $a = 1$ and $b = -3$ :

$f''(x) = 6x + 18$ Setting this greater than zero:

$6x + 18 > 0$ Then solving gives:

$6x > -18$ $x > -3$ Using the graph, we confirm $f''(x) > 0$ for values of $x$ greater than -3.

Step 4

For which values of $c$ will the graph of $p(x) = f(x) + t$ have two distinct positive roots and one negative root?

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Answer

To ensure the graph has two distinct positive roots and one negative root, the function needs to cross the x-axis three times. The condition on $c$ is related to the vertical shift of $f(x)$ . Setting the discriminant of $f(x)$ to be positive while ensuring that the graph remaining intersecting the x-axis in the required manner leads to:

$-32 < c < 5$ This indicates that as $c$ varies within this range, the equation will fulfill the criteria of having the necessary distinct roots.

The graph of $f(x) = ax^3 + bx^2 + cx - 5$ is drawn below - NSC Mathematics - Question 9 - 2024 - Paper 1

Worked Solution & Example Answer:The graph of $f(x) = ax^3 + bx^2 + cx - 5$ is drawn below - NSC Mathematics - Question 9 - 2024 - Paper 1

Show that $a = 1$, $b = -3$ and $c = -9$.

Calculate the value of $x$ for which $f$ has a local minimum value.

Use the graph to determine the values of $x$ for which $f''(x) > 0$.

For which values of $c$ will the graph of $p(x) = f(x) + t$ have two distinct positive roots and one negative root?

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