Leaving Cert Mathematics Algebra: The diagram shows the graph of a

The diagram shows the graph of a function $f(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2022

Question b

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The diagram shows the graph of a function $f(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{Z}$. Three regions on the diagram are marked K, L, and N. Each of these ... show full transcript

Worked Solution & Example Answer:The diagram shows the graph of a function $f(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2022

Step 1

The area of region K is 538 square units. Use integration of $f(x)$ to show that:

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Answer

To find the area of region K, we can use integration. The area under curve $f(x)$ from $x=0$ to $x=2$ can be computed as:

$\text{Area} = \int_0^2 f(x) \, dx = \int_0^2 (ax^2 + bx + c) \, dx$

Calculating the integral:

$= \left[ \frac{a}{3} x^3 + \frac{b}{2} x^2 + cx \right]_0^2$ $= \frac{a}{3} (2^3) + \frac{b}{2} (2^2) + c(2) - 0$ $= \frac{8a}{3} + 2b + 2c$

Setting this equal to 538 gives:

$\frac{8a}{3} + 2b + 2c = 538$

Multiplying the entire equation by 3 to eliminate the fraction:

$8a + 6b + 6c = 1614$

Now simplifying this with the equation in mark scheme yields:

$4a + 3b + 3c = 807$ .

Step 2

The areas of the three regions K, L, and N give the following three equations (including the equation from part (b)(i)):

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Answer

To solve for a, b, and c, we use the three equations derived from the areas of the regions K, L, and N:

$4a + 3b + 3c = 807 \, \text{ (from part (b)(i))}$
$28a + 9b + 3c = 879$
$76a + 15b + 3c = 663$

Now we will solve these equations step by step:

Subtract the first equation from the second equation: $(28a + 9b + 3c) - (4a + 3b + 3c) = 879 - 807$ This leads to: $24a + 6b = 72 \, \Rightarrow \; 4a + b = 12 \, \text{(Eq. 1)}$
Now subtract the first equation from the third: $(76a + 15b + 3c) - (4a + 3b + 3c) = 663 - 807$ Leading to: $72a + 12b = -144 \, \Rightarrow \; 6a + b = -12 \, \text{(Eq. 2)}$

With the two equations (Eq. 1) and (Eq. 2):

Set the equations in terms of $b$ :
- From Eq. 1: $b = 12 - 4a$
- Substitute this into Eq. 2: $6a + (12 - 4a) = -12$ Simplifying yields: $2a = -24 \, \Rightarrow \; a = -12$
Substitute $a = -12$ back into Eq. 1: $4(-12) + b = 12 \Rightarrow -48 + b = 12 \Rightarrow b = 60$
Finally, substitute $a = -12$ and $b = 60$ back into one of the original equations to find c: $4(-12) + 3(60) + 3c = 807$ $-48 + 180 + 3c = 807 \Rightarrow 3c = 675 \Rightarrow c = 225$

Thus, we have:

$a = -12$
$b = 60$
$c = 225$ .

The diagram shows the graph of a function $f(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2022

Worked Solution & Example Answer:The diagram shows the graph of a function $f(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{Z}$ - Leaving Cert Mathematics - Question b - 2022

The area of region K is 538 square units. Use integration of $f(x)$ to show that:

The areas of the three regions K, L, and N give the following three equations (including the equation from part (b)(i)):

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