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 $x$ 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 $c = -9$. 9.2 Ca... 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 values of $a$ , $b$ , and $c$ , we substitute the x-intercept values from points E and G into the function:

At $x = -1$ , $f(-1) = 0$ :

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

Rearranging gives:

ightarrow b - c - a = 5 ag{1}$$

At $x = 5$ , $f(5) = 0$ :

$f(5) = a(5)^3 + b(5)^2 + c(5) - 5 = 125a + 25b + 5c - 5 = 0$

Rearranging gives: $125a + 25b + 5c = 5 ag{2}$
Substituting $a = 1$ into equations (1) and (2):

From (1):

ightarrow b - c = 6 ag{3}$$

From (2):

ightarrow 25b + 5c = -120$$ Dividing throughout by 5 gives: $$5b + c = -24 ag{4}$$ Now, solving equations (3) and (4) together gives: Substituting (3) into (4), we find: $$5b + (b - 6) = -24$$ $$6b - 6 = -24 ightarrow 6b = -18 ightarrow b = -3$$ Now substituting back to find $c$: $$b - c = 6 ightarrow -3 - c = 6 ightarrow c = -9$$ Hence, $a = 1$, $b = -3$, 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 x-values for local minimum, we need to compute the first derivative and set it to zero:

The first derivative of $f(x)$ is: $f'(x) = 3ax^2 + 2bx + c$
With $a = 1$ , $b = -3$ , and $c = -9$ , the derivative becomes: $f'(x) = 3x^2 - 6x - 9$
Setting $f'(x) = 0$ gives:

ightarrow x^2 - 2x - 3 = 0 $4. Factoring:$ (x - 3)(x + 1) = 0 $So, $x = 3$ or $x = -1$ 5. To determine if these points correspond to a minimum, we compute the second derivative:$ f''(x) = 6x - 6$$ Evaluating at $x = 3$ :

ightarrow ext{Local minimum at } x = 3.$$ At $x = -1$: $$f''(-1) = 6(-1) - 6 = -12 < 0 ightarrow ext{Local maximum at } x = -1.$$

Step 3

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

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Answer

From the second derivative: $f''(x) = 6x - 6$ We set this greater than zero:

ightarrow x > 1$$ Hence, $f''(x) > 0$ for $x > 1$. This indicates that the function is concave up in that interval.

Step 4

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

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Answer

For the cubic $p(x) = f(x) + t$ to have two distinct positive roots and one negative root, the following conditions must be fulfilled:

$t$ must be adjusted so that the function cuts the x-axis three times.
Since $f(-1) = 0$ and $f(5) = 0$ , we need to ensure that $p(t)$ shifts the graph up or down in most scenarios by less than 32 units, as indicated by the preceding behavior of the function, to achieve the desired root patterns.
- Thus, we deduce:

ightarrow -32 < t < 32$$

Hence, the suitable range for $t$ is $-32 < t < 32$ .

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

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