The function is $g(x) = x + 3, orall x ext{ in } ext{R}$. The points $A(1, k)$ and $B$ are the points of intersection of $f$ and $g$. Find the co-ordinates of $A$ and of $B$. The points $(0, 0)$ and $(5, 0)$ are on the graph of $f$. (i) Draw the quadrilateral $OCBA$ on the diagram above. (ii) Find the area of the quadrilateral $OCBA$.

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The function is $g(x) = x + 3, orall x ext{ in } ext{R}$ - Leaving Cert Mathematics - Question 5 - 2015

Question 5

The function is $g(x) = x + 3, orall x ext{ in } ext{R}$. The points $A(1, k)$ and $B$ are the points of intersection of $f$ and $g$. Find the co-ordinates of... show full transcript

Worked Solution & Example Answer:The function is $g(x) = x + 3, orall x ext{ in } ext{R}$ - Leaving Cert Mathematics - Question 5 - 2015

Step 1

Find the Intersection Points A and B

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Answer

To find the intersection points of the functions $f(x) = 5x - x^2$ and $g(x) = x + 3$ , we set them equal: $5x - x^2 = x + 3$ Rearranging gives:

-x^2 + 4x - 3 = 0 \\

We can multiply through by -1: $x^2 - 4x + 3 = 0$ Factoring: $(x - 1)(x - 3) = 0$ Thus, $x = 1$ and $x = 3$ . Substituting these x-values back into $g(x)$ to find the y-coordinates:

For $A(1, k)$ :

A(1, 4)$$ - For $B$: $$g(3) = 3 + 3 = 6 \\ B(3, 6)$$

Step 2

Draw the quadrilateral OCBA

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Answer

Using the points $O(0, 0)$ , $C(5, 0)$ , $B(3, 6)$ , and $A(1, 4)$ , draw the quadrilateral $OCBA$ on the provided diagram by connecting these points in sequence.

Step 3

Find the area of the quadrilateral OCBA

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Answer

To calculate the area of quadrilateral $OCBA$ , we can use the formula: $ext{Area} = | riangle ODA| + | riangle DEB| + | riangle ACB|$ Let’s identify each area:

For triangle $OAD$ : $A(1, 4), O(0, 0), D(0, 6)$ with base $1$ and height $4$ : $ext{Area}_{OAD} = rac{1}{2} imes ext{base} imes ext{height} = rac{1}{2} imes 1 imes 4 = 2$
For triangle $DEB$ : $D(0, 6), E(5, 0), B(3, 6)$ , the area calculation yields: $ext{Area}_{DEB} = rac{1}{2} imes (5)(6) = 15$
For triangle $ACB$ : Using the coordinates: $ext{Area}_{ACB} = rac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Plugging in gives: $rac{1}{2}|0(0 - 6) + 5(6 - 4) + 3(4 - 0)| = rac{1}{2}|0 + 10 + 12| = 11$ Total area: $ext{Area}_{OCBA} = 2 + 15 + 11 = 18 ext{ square units}$

The function is $g(x) = x + 3, orall x ext{ in } ext{R}$ - Leaving Cert Mathematics - Question 5 - 2015

Worked Solution & Example Answer:The function is $g(x) = x + 3, orall x ext{ in } ext{R}$ - Leaving Cert Mathematics - Question 5 - 2015

Find the Intersection Points A and B

Draw the quadrilateral OCBA

Find the area of the quadrilateral OCBA

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