The table in Part (a)(ii) below shows some of the values of the function: $$h(x) = 0.001x^3 - 0.12x^2 + px + 5, \, x \in \mathbb{R}$$ in the domain $$0 \leq x \leq 75$$. (i) Use $$h(10) = 30$$ to show that $$p = 3.6$$. (ii) Complete the table below and hence draw the graph of $$h(x)$$ in the domain $$0 \leq x \leq 75$$ on the grid below. | x | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 75 | |-----|----|----|----|----|----|----|----|----|----| | h(x)| 30 | | | | | | | | |

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The table in Part (a)(ii) below shows some of the values of the function: $$h(x) = 0.001x^3 - 0.12x^2 + px + 5, \, x \in \mathbb{R}$$ in the domain $$0 \leq x \leq 75$$ - Leaving Cert Mathematics - Question 8(a) - 2021

Question 8(a)

$The-table-in-Part-(a)(ii)-below-shows-some-of-the-values-of-the-function:--$$h(x)-=-0.001x^3---0.12x^2-+-px-+-5,-\,-x-\in-\mathbb{R}$$--in-the-domain-$$0-\leq-x-\leq-75$$-Leaving Cert Mathematics-Question 8(a)-2021.png$

The table in Part (a)(ii) below shows some of the values of the function: $$h(x) = 0.001x^3 - 0.12x^2 + px + 5, \, x \in \mathbb{R}$$ in the domain $$0 \leq x \leq... show full transcript

Worked Solution & Example Answer:The table in Part (a)(ii) below shows some of the values of the function: $$h(x) = 0.001x^3 - 0.12x^2 + px + 5, \, x \in \mathbb{R}$$ in the domain $$0 \leq x \leq 75$$ - Leaving Cert Mathematics - Question 8(a) - 2021

Step 1

Use $$h(10) = 30$$ to show that $$p = 3.6$$

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Answer

To find the value of $p$ , we substitute $x = 10$ into the function:

$h(10) = 0.001(10)^3 - 0.12(10)^2 + p(10) + 5$

Calculating each term:

$0.001(10)^3 = 0.001 \cdot 1000 = 1$
$-0.12(10)^2 = -0.12 \cdot 100 = -12$
The expression becomes:

$h(10) = 1 - 12 + 10p + 5 = 30$

Combining the constants:

$-6 + 10p = 30$

Adding 6 to both sides gives:

$10p = 36$

Now, dividing by 10:

$p = 3.6$

Step 2

Complete the table below and hence draw the graph of $$h(x)$$ in the domain $$0 \leq x \leq 75$$

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Answer

To complete the table, we will calculate the values of $h(x)$ for the missing inputs:

For $x = 20$ : $h(20) = 0.001(20)^3 - 0.12(20)^2 + 3.6(20) + 5$ Compute: $h(20) = 0.001(8000) - 0.12(400) + 72 + 5 = 8 - 48 + 72 + 5 = 37$
For $x = 30$ : $h(30) = 0.001(30)^3 - 0.12(30)^2 + 3.6(30) + 5$ Compute: $h(30) = 0.001(27000) - 0.12(900) + 108 + 5 = 27 - 108 + 108 + 5 = 32$
For $x = 40$ : $h(40) = 0.001(64,000) - 0.12(1600) + 144 + 5 = 64 - 192 + 144 + 5 = 21$
For $x = 50$ : $h(50) = 0.001(125,000) - 0.12(2500) + 180 + 5 = 125 - 300 + 180 + 5 = 10$
For $x = 60$ : $h(60) = 0.001(216,000) - 0.12(3600) + 216 + 5 = 216 - 432 + 216 + 5 = 5$
For $x = 70$ : $h(70) = 0.001(343,000) - 0.12(4900) + 252 + 5 = 343 - 588 + 252 + 5 = 12$
For $x = 75$ : $h(75) = 0.001(421875) - 0.12(5625) + 270 + 5 = 421.875 - 675 + 270 + 5 = 21.875$

The completed table is:

x	0	10	20	30	40	50	60	70	75
h(x)	30	30	37	32	21	10	5	12	21.875

To draw the graph, plot all these points on the grid, ensuring the appropriate connections are made between points.

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The table in Part (a)(ii) below shows some of the values of the function: $$h(x) = 0.001x^3 - 0.12x^2 + px + 5, \, x \in \mathbb{R}$$ in the domain $$0 \leq x \leq 75$$ - Leaving Cert Mathematics - Question 8(a) - 2021

Worked Solution & Example Answer:The table in Part (a)(ii) below shows some of the values of the function: $$h(x) = 0.001x^3 - 0.12x^2 + px + 5, \, x \in \mathbb{R}$$ in the domain $$0 \leq x \leq 75$$ - Leaving Cert Mathematics - Question 8(a) - 2021

Use $$h(10) = 30$$ to show that $$p = 3.6$$

Complete the table below and hence draw the graph of $$h(x)$$ in the domain $$0 \leq x \leq 75$$

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