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The graph of a cubic function $f(x)$ cuts the x-axis at $x = -3$, $x = -1$ and $x = 2$, and the y-axis at $(0, -6)$, as shown - Leaving Cert Mathematics - Question 1 - 2014

Question 1

The graph of a cubic function $f(x)$ cuts the x-axis at $x = -3$, $x = -1$ and $x = 2$, and the y-axis at $(0, -6)$, as shown. Verify that $f(x)$ can be written as ... show full transcript

Worked Solution & Example Answer:The graph of a cubic function $f(x)$ cuts the x-axis at $x = -3$, $x = -1$ and $x = 2$, and the y-axis at $(0, -6)$, as shown - Leaving Cert Mathematics - Question 1 - 2014

Step 1

Verify that $f(x)$ can be written as $f(x) = x^3 + 2x^2 - 5x - 6$

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Answer

To verify that the function can be expressed in the desired form, we will utilize the roots $x = -3$ , $x = -1$ , and $x = 2$ .

The cubic function can be factored as follows:

$f(x) = k(x + 3)(x + 1)(x - 2)$

Here, we substitute each root to determine the leading coefficient $k$ :

For $x = -3$ : $f(-3) = k(0) = 0$
For $x = -1$ : $f(-1) = k(1) = 0$
For $x = 2$ : $f(2) = k((-1)(-2)(0)) = 0$

Next, we can expand the expression:

$f(x) = k(x + 3)(x + 1)(x - 2)$

Expanding this:

First, multiply $(x + 3)(x + 1)$ :

$(x + 3)(x + 1) = x^2 + 4x + 3$

Next, multiply that by $(x - 2)$ :

$(x^2 + 4x + 3)(x - 2) = x^3 + 2x^2 - 5x - 6$

This verifies the given function. Thus, $k = 1$ leads us to the final expression.

Therefore, we have verified that:

$f(x) = x^3 + 2x^2 - 5x - 6$

Step 2

The graph of the function $g(x) = -2x - 6$ intersects the graph of the function $f(x)$ above. Let $f(x) = g(x)$ and solve the resulting equation to find the co-ordinates of the points where the graphs of $f(x)$ and $g(x)$ intersect.

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Answer

To find the intersection points, set the functions equal to each other:

$x^3 + 2x^2 - 5x - 6 = -2x - 6$

Rearranging gives:

$x^3 + 2x^2 - 5x + 2x = 0$

This simplifies to:

$x^3 + 2x^2 - 3x = 0$

Now, factor out $x$ :

$x(x^2 + 2x - 3) = 0$

This gives:

$x = 0$
Now, solve $x^2 + 2x - 3 = 0$ using the quadratic formula:

oot{b^2 - 4ac}}{2a}$$ Here $a = 1$, $b = 2$, and $c = -3$: $$x = rac{-2 ext{ ± } oot{2^2 - 4(1)(-3)}}{2(1)}$$ This computes to: $$x = rac{-2 ext{ ± } 4}{2}$$ Thus yielding: - $x = 1$ (positive) - $x = -3$ (negative) Now, substitute the $x$ values into either function to find the corresponding $y$ values: - For $x = 0$: $$g(0) = -2(0) - 6 = -6 ightarrow (0, -6)$$ - For $x = 1$: $$g(1) = -2(1) - 6 = -8 ightarrow (1, -8)$$ - For $x = -3$: $$g(-3) = -2(-3) - 6 = 0 ightarrow (-3, 0)$$ Thus, the points of intersection are: $$(-3, 0), (0, -6), (1, -8)$$

Step 3

Draw the graph of the function $g(x) = -2x - 6$ on the diagram above.

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Answer

To draw the graph of the function $g(x) = -2x - 6$ , we first find its intercepts:

Y-Intercept: When $x = 0$ :

$g(0) = -2(0) - 6 = -6$

This gives the point $(0, -6)$ .
X-Intercept: Set $g(x) = 0$ :

$0 = -2x - 6$

Thus, solving for $x$ gives:

ightarrow x = -3$$

This gives us the point $(-3, 0)$ .

Using these points, plot $(0, -6)$ and $(-3, 0)$ on the graph above. Then, draw a straight line connecting these points, indicating the linear nature of $g(x) = -2x - 6$ .

The graph of a cubic function $f(x)$ cuts the x-axis at $x = -3$, $x = -1$ and $x = 2$, and the y-axis at $(0, -6)$, as shown - Leaving Cert Mathematics - Question 1 - 2014

Worked Solution & Example Answer:The graph of a cubic function $f(x)$ cuts the x-axis at $x = -3$, $x = -1$ and $x = 2$, and the y-axis at $(0, -6)$, as shown - Leaving Cert Mathematics - Question 1 - 2014

Verify that $f(x)$ can be written as $f(x) = x^3 + 2x^2 - 5x - 6$

The graph of the function $g(x) = -2x - 6$ intersects the graph of the function $f(x)$ above. Let $f(x) = g(x)$ and solve the resulting equation to find the co-ordinates of the points where the graphs of $f(x)$ and $g(x)$ intersect.

Draw the graph of the function $g(x) = -2x - 6$ on the diagram above.

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