Junior Cycle Mathematics ALGEBRA - Expressions: The function $h(x)$ below gives

The function $h(x)$ below gives the approximate height of the water at Howth Harbour on a particular day, from 12 noon to 5 p.m., where $h(x) = 10x^2 - 50x + 130$, where $h(x)$ is the height of the water in centimeters, and $x$ is the time in hours after 12 noon - Junior Cycle Mathematics - Question 14 - 2016

Question 14

The-function-$h(x)$-below-gives-the-approximate-height-of-the-water-at-Howth-Harbour-on-a-particular-day,-from-12-noon-to-5-p.m.,--where-$h(x)-=-10x^2---50x-+-130$,---where-$h(x)$-is-the-height-of-the-water-in-centimeters,-and-$x$-is-the-time-in-hours-after-12-noon-Junior Cycle Mathematics-Question 14-2016.png

The function $h(x)$ below gives the approximate height of the water at Howth Harbour on a particular day, from 12 noon to 5 p.m., where $h(x) = 10x^2 - 50x + 130$, ... show full transcript

Worked Solution & Example Answer:The function $h(x)$ below gives the approximate height of the water at Howth Harbour on a particular day, from 12 noon to 5 p.m., where $h(x) = 10x^2 - 50x + 130$, where $h(x)$ is the height of the water in centimeters, and $x$ is the time in hours after 12 noon - Junior Cycle Mathematics - Question 14 - 2016

Step 1

Draw the graph of the function $h(x)$

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Answer

To draw the graph of the function $h(x) = 10x^2 - 50x + 130$ , I will first find its vertex, which corresponds to the lowest point of the parabola. The vertex can be found using the formula:

$x = -\frac{b}{2a} = -\frac{-50}{2 \cdot 10} = 2.5$

Next, substitute $x = 2.5$ back into the function to find the corresponding $h(2.5)$ :

$h(2.5) = 10(2.5)^2 - 50(2.5) + 130 = 10(6.25) - 125 + 130 = 62.5 + 5 = 67.5$

So the lowest point is at $(2.5, 67.5)$ . At $x = 0$ (12 noon), the height is:

$h(0) = 10(0)^2 - 50(0) + 130 = 130$

At $x = 5$ (5 p.m.),

$h(5) = 10(5)^2 - 50(5) + 130 = 10(25) - 250 + 130 = -25 + 130 = 105$

Now plot these points and connect them smoothly to draw the parabola.

Step 2

Find the height of the water at 12 noon.

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Answer

At 12 noon ( $x = 0$ ), the height of the water is:

$h(0) = 130 \text{ cm}$

Step 3

Estimate the height of the water at its lowest point.

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Answer

The height of the water at its lowest point (vertex of the parabola) is approximately:

$67.5 \text{ cm}$

Step 4

After 12 noon, how long did it take before the water was at its lowest point?

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Answer

It took:

$2.5 \text{ hours}$

before the water reached its lowest point.

Step 5

Find the value of $c$.

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Answer

From the information given, at the lowest point, $g(3) = 0$ . At $(0, 180)$ ,

$c = 180$ .

Step 6

Hence, or otherwise, find the value of $a$ and the value of $b$.

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Answer

Using the equation of the quadratic, since the vertex occurs at $x = 3$ , we also have:

$g(x) = a(x - 3)^2 + 180$

By substituting $(6, 0)$ into the equation:

$0 = a(3)^2 + 180$

Solving this gives:

$a = 20$

Next, by substituting into the original function:

g(x) = 20x^2 - 120x + 180$$ while simplifying gives:

$$b = -120.$

Draw the graph of the function $h(x)$

Find the height of the water at 12 noon.

Estimate the height of the water at its lowest point.

After 12 noon, how long did it take before the water was at its lowest point?

Find the value of $c$.

Hence, or otherwise, find the value of $a$ and the value of $b$.

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