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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 $f(x)$ c... 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

The y-intercept of the function f(x)f(x) can be found by substituting x=0x = 0 into the equation:

f(0)=03+4(0)27(0)10=10f(0) = 0^3 + 4(0)^2 - 7(0) - 10 = -10

Thus, the y-intercept is 10-10.

Step 2

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

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Answer

To show that 22 is a root, we will substitute x=2x = 2 into f(x)f(x):

f(2)=23+4(2)27(2)10=8+161410=0f(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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Since 22 is a root, we can factor f(x)f(x) using synthetic division:

  1. Perform synthetic division of f(x)f(x) by (x2)(x - 2):

f(x)=(x2)(x2+6x+5)f(x) = (x - 2)(x^2 + 6x + 5)

  1. The quadratic x2+6x+5x^2 + 6x + 5 can be further factored as:

(x+1)(x+5)(x + 1)(x + 5)

  1. Therefore, the complete factorisation 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

To draw the graph:

  1. Plot the y-intercept at (0,10)(0, -10).
  2. Identify the x-intercepts by solving f(x)=0f(x) = 0, which occur at x=5,1,2x = -5, -1, 2.
  3. Mark the turning points at (0.7;12.6)(0.7; 12.6) and (3.4;20.8)(-3.4; 20.8).
  4. Sketch the curve that connects these points, being mindful of the behavior of cubic functions around turning points.

Step 5

8.5.1 f'(x) < 0

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To determine where f(x)<0f'(x) < 0, analyze the graph of ff. It shows that the function is decreasing between the turning points, indicating where the derivative is negative.

Step 6

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

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Answer

The gradient of the tangent is minimum at the point where the derivative is zero. This occurs at the turning points of the graph, which can be identified at x=3.4x = -3.4 and x=0.7x = 0.7.

Step 7

8.5.3 f'(x) . f''(x) ≤ 0

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To satisfy f'(x) ullet f''(x) ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ }:

  1. Identify intervals where the product of the derivative and second derivative is less than or equal to zero, typically at critical points and boundaries of increasing/decreasing intervals.

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