A wall plaque is to be made to commemorate the 150th anniversary of the publication of "Alice's Adventures in Wonderland". The edges of the wall plaque can be modelled by parts of the graphs of four quadratic functions as shown in the sketch. - $f(x) = \frac{1}{4}x^2 + \frac{1}{3}x + 3$ - $g(x) = -\frac{1}{3}x^2 + \frac{5}{3}$ - $h(x) = \frac{1}{2}x^2 - 2x + 1$ - $k(x) = -\frac{1}{6}x^2 + 3$ (a) Find the x-coordinate of the point of intersection of the graphs with equations $y = f(x)$ and $y = g(x)$. The graphs of the functions $f(x)$ and $h(x)$ intersect on the y-axis. The plaque has a vertical line of symmetry. (b) Calculate the area of the wall plaque.

Scottish Highers Maths Integration: A wall plaque is to be made to c

A wall plaque is to be made to commemorate the 150th anniversary of the publication of "Alice's Adventures in Wonderland" - Scottish Highers Maths - Question 4 - 2015

Question 4

A-wall-plaque-is-to-be-made-to-commemorate-the-150th-anniversary-of-the-publication-of-"Alice's-Adventures-in-Wonderland"-Scottish Highers Maths-Question 4-2015.png

A wall plaque is to be made to commemorate the 150th anniversary of the publication of "Alice's Adventures in Wonderland". The edges of the wall plaque can be model... show full transcript

Worked Solution & Example Answer:A wall plaque is to be made to commemorate the 150th anniversary of the publication of "Alice's Adventures in Wonderland" - Scottish Highers Maths - Question 4 - 2015

Step 1

Find the x-coordinate of the point of intersection of the graphs with equations $y = f(x)$ and $y = g(x)$

96%

114 rated

Answer

To find the point of intersection, we equate the two functions:

$f(x) = g(x)$

This leads to:

$\frac{1}{4}x^2 + \frac{1}{3}x + 3 = -\frac{1}{3}x^2 + \frac{5}{3}$

Bringing all terms to one side, we have:

$\frac{1}{4}x^2 + \frac{1}{3}x + 3 + \frac{1}{3}x^2 - \frac{5}{3} = 0$

Combining the x^2 terms:

$\left(\frac{1}{4} + \frac{1}{3}\right)x^2 + \frac{1}{3}x + \left(3 - \frac{5}{3}\right) = 0$

Calculating terms gives:

$\left(\frac{3}{12} + \frac{4}{12}\right)x^2 + \frac{1}{3}x + \frac{4}{3} = 0$

Which simplifies to:

$\frac{7}{12}x^2 + \frac{1}{3}x + \frac{4}{3} = 0$

To solve for x, multiplying through by 12 to eliminate fractions yields:

$7x^2 + 4x + 16 = 0$

Now using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 7$ , $b = 4$ , and $c = 16$ . Thus,

$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 7 \cdot 16}}{2 \cdot 7}$

Calculating the discriminant:

$16 - 448 = -432$

Since the discriminant is negative, there are no real solutions, indicating the graphs do not intersect.

Step 2

Calculate the area of the wall plaque

99%

104 rated

Answer

To calculate the area of the wall plaque, we will find the area between the curves $f(x)$ and $h(x)$ , as the plaque has points of intersection along the y-axis.

The area $A$ can be obtained by integrating the difference of the two functions:

$A = \int_{a}^{b} \left( f(x) - h(x) \right) dx$

where $a$ and $b$ are the limits of integration defined by the intersections.

Thus, we first need the intersection points:

Setting $f(x) = h(x)$ leads to:

$\frac{1}{4}x^2 + \frac{1}{3}x + 3 = \frac{1}{2}x^2 - 2x + 1$

Rearranging gives:

$\left(\frac{1}{4} - \frac{1}{2}\right)x^2 + \left(\frac{1}{3} + 2\right)x + 3 - 1 = 0$

which simplifies to:

$-\frac{1}{4}x^2 + \frac{7}{3}x + 2 = 0$

Multiplying through by $-4$ :

$x^2 - \frac{28}{3}x - 8 = 0$

We can find the solutions using the quadratic formula again. The limits of integration, once established, can be substituted into the area integral.

Finally, summing the output will yield the area of the wall plaque.

A wall plaque is to be made to commemorate the 150th anniversary of the publication of "Alice's Adventures in Wonderland" - Scottish Highers Maths - Question 4 - 2015

Worked Solution & Example Answer:A wall plaque is to be made to commemorate the 150th anniversary of the publication of "Alice's Adventures in Wonderland" - Scottish Highers Maths - Question 4 - 2015

Find the x-coordinate of the point of intersection of the graphs with equations $y = f(x)$ and $y = g(x)$

Calculate the area of the wall plaque

Join the Scottish Highers students using SimpleStudy...