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Given: $$f(x) = x^3 + 4x^2 - 7x - 10$$ 8.1 Write down the y-intercept of $f$ - NSC Mathematics - Question 8 - 2023 - Paper 1

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Given:--$$f(x)-=-x^3-+-4x^2---7x---10$$--8.1-Write-down-the-y-intercept-of-$f$-NSC Mathematics-Question 8-2023-Paper 1.png

Given: $$f(x) = x^3 + 4x^2 - 7x - 10$$ 8.1 Write down the y-intercept of $f$. 8.2 Show that $2$ is a root of the equation $f(x) = 0$. 8.3 Hence, factorise $f(x)$... show full transcript

Worked Solution & Example Answer:Given: $$f(x) = x^3 + 4x^2 - 7x - 10$$ 8.1 Write down the y-intercept of $f$ - NSC Mathematics - Question 8 - 2023 - Paper 1

Step 1

8.1 Write down the y-intercept of $f$.

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Answer

To find the y-intercept of the function f(x)f(x), we evaluate it at x=0x = 0:

f(0)=03+4(0)27(0)10=10.f(0) = 0^3 + 4(0)^2 - 7(0) - 10 = -10.
Thus, the y-intercept is (0,10)(0, -10).

Step 2

8.2 Show that $2$ is a root of the equation $f(x) = 0$.

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To confirm that 22 is a root, we compute:

f(2)=(2)3+4(2)27(2)10=8+161410=0.f(2) = (2)^3 + 4(2)^2 - 7(2) - 10 = 8 + 16 - 14 - 10 = 0.
Since f(2)=0f(2) = 0, 22 is indeed a root of the equation.

Step 3

8.3 Hence, factorise $f(x)$ completely.

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We already established that (x2)(x - 2) is a root. We can use polynomial long division to divide f(x)f(x) by (x2)(x - 2):

f(x)=(x2)(x2+6x+5).f(x) = (x - 2)(x^2 + 6x + 5).
Now, we factor x2+6x+5x^2 + 6x + 5:

x2+6x+5=(x+1)(x+5).x^2 + 6x + 5 = (x + 1)(x + 5).
Thus, the complete factorization is:

f(x)=(x2)(x+1)(x+5).f(x) = (x - 2)(x + 1)(x + 5).

Step 4

8.4 If it is further given that the coordinates of the turning points are approximately at $(0.7; 12.6)$ and $(-3.4; 20.8)$, draw a sketch graph of $f$ and label all intercepts and turning points.

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Answer

For the sketch of the graph, plot the y-intercept (0,10)(0, -10), the turning points (0.7,12.6)(0.7, 12.6) and (3.4,20.8)(-3.4, 20.8), and the x-intercepts found from the factorization: (2,0)(2, 0), (1,0)(-1, 0), and (5,0)(-5, 0).
Connect these points smoothly to illustrate the shape of the cubic function, ensuring to maintain the characteristics of turning points.

Step 5

8.5.1 $f'(x) < 0$

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From the graph analysis, identify the sections where the curve is decreasing. This occurs when x is between the turning points, specifically for intervals approximately (3.4,0.7)(-3.4, 0.7).

Step 6

8.5.2 The gradient of a tangent to $f$ will be a minimum.

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To find where the gradient is at a minimum point, calculate the derivative f(x)f'(x) and analyze critical points.
The minimum gradient is reached at the turning point (3.4,20.8)(-3.4, 20.8), indicating that the gradient transitions from negative to positive.

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