The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below. A and B are the turning points of $f$. $C(2; 4)$ is a point on $f$. 9.1 Determine the coordinates of A and B. 9.2 For which values of $x$ will $f$ be concave up? 9.3 Determine the equation of the tangent to $f$ at $C(2; 4)$.

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The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below - NSC Mathematics - Question 9 - 2021 - Paper 1

Question 9

The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below. A and B are the turning points of $f$. $C(2; 4)$ is a point on $f$. 9.1 Determine the coordinates of A a... show full transcript

Worked Solution & Example Answer:The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below - NSC Mathematics - Question 9 - 2021 - Paper 1

Step 1

Determine the coordinates of A and B.

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Answer

To find the turning points of the function, we need to compute the first derivative:

$f'(x) = 6x^2 + 6x - 12$

Setting the first derivative to zero:

$6x^2 + 6x - 12 = 0$

Dividing the entire equation by 6:

$x^2 + x - 2 = 0$

Factoring gives us:

$(x - 1)(x + 2) = 0$

This yields the solutions:

$x = 2 \quad \text{or} \quad x = -1$

Now substituting back into the original function to find the corresponding $y$ coordinates:

For $x = 2$ :

$f(2) = 2(2)^3 + 3(2)^2 - 12(2) = 2(8) + 3(4) - 24 = 16 + 12 - 24 = 4$

So one turning point is $A(2; 4)$ .

For $x = -1$ :

$f(-1) = 2(-1)^3 + 3(-1)^2 - 12(-1) = 2(-1) + 3(1) + 12 = -2 + 3 + 12 = 13$

So the other turning point is $B(-1; 13)$ .

Step 2

For which values of x will f be concave up?

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Answer

To determine the concavity of $f$ , we need to compute the second derivative:

$f''(x) = 12x + 6$

Setting the second derivative greater than zero:

$12x + 6 > 0$

Solving for $x$ :

$12x > -6 \implies x > -\frac{1}{2}$

Thus, $f$ is concave up for $x > -\frac{1}{2}$ .

Step 3

Determine the equation of the tangent to f at C(2; 4).

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Answer

The equation of the tangent line can be formed using the point-slope form:

$y - y_1 = m(x - x_1)$

Where $(x_1, y_1)$ is the point on the curve, which is $C(2, 4)$ .

First, we need to find the derivative at $x = 2$ to get the slope $m$ :

$f'(2) = 6(2)^2 + 6(2) - 12 = 6(4) + 12 - 12 = 24$

Now substituting the values into the point-slope form:

$y - 4 = 24(x - 2)$

This simplifies to:

$y - 4 = 24x - 48 \implies y = 24x - 44$

Thus, the equation of the tangent is $y = 24x - 44$ .

The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below - NSC Mathematics - Question 9 - 2021 - Paper 1

Worked Solution & Example Answer:The graph of $f(x) = 2x^3 + 3x^2 - 12x$ is sketched below - NSC Mathematics - Question 9 - 2021 - Paper 1

Determine the coordinates of A and B.

For which values of x will f be concave up?

Determine the equation of the tangent to f at C(2; 4).

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