Question 7
(a) Complete the table of values for
$y = 0.5^{\left(\frac{x}{3}+1\right)} + 2$
giving each value to 2 decimal places where appropriate.
$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline\\x&-6&-5&-4&-3&-2&-1&0\\ \hline y&4&3.59&3.26&&&&2.5\\ \hline\end{array}$$(2)
(b) On the grid opposite, draw the graph of
$y = 0.5^{\left(\frac{x}{3}+1\right)} + 2$
for $-6 \leqslant x \leqslant 0$
(2)
(c) By drawing a suitable straight line on the grid, obtain an estimate, to one decimal place, of the root of the equation
$\log_2(2x+2)^3 + x + 3 = 0$
in the interval $-6 \leqslant x \leqslant 0$
(6)
Step (a) Complete the table of values for
The given function is
$y = 0.5^{\left(\frac{x}{3}+1\right)} + 2$
We are tasked with completing the table for values of $x$ from $-6$ to $0$. Using the formula for $y$, we calculate the corresponding values of $y$ for each $x$:
For $x = -6$:
$y = 0.5^{\left(\frac{-6}{3}+1\right)} + 2 = 0.5^{(-2+1)} + 2 = 0.5^{-1} + 2 = 2 + 2 = 4$
For $x = -5$:
$y = 0.5^{\left(\frac{-5}{3}+1\right)} + 2 = 0.5^{(-1.6667+1)} + 2 = 0.5^{-0.6667} + 2 \approx 0.792 + 2 = 3.59$
For $x = -4$:
$y = 0.5^{\left(\frac{-4}{3}+1\right)} + 2 = 0.5^{(-1.3333+1)} + 2 = 0.5^{-0.3333} + 2 \approx 0.7937 + 2 = 3.26$
For $x = -3$:
$y = 0.5^{\left(\frac{-3}{3}+1\right)} + 2 = 0.5^{(0)} + 2 = 1 + 2 = 3$
For $x = -2$:
$y = 0.5^{\left(\frac{-2}{3}+1\right)} + 2 = 0.5^{(0.3333)} + 2 \approx 0.7937 + 2 = 2.79$
For $x = -1$:
$y = 0.5^{\left(\frac{-1}{3}+1\right)} + 2 = 0.5^{(0.6667)} + 2 \approx 0.6299 + 2 = 2.6299$
For $x = 0$:
$y = 0.5^{\left(\frac{0}{3}+1\right)} + 2 = 0.5^{(1)} + 2 = 0.5 + 2 = 2.5$
Thus, the completed table of values is:
$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline\\x&-6&-5&-4&-3&-2&-1&0\\ \hline y&4&3.59&3.26&3&2.79&2.63&2.5\\ \hline\end{array}$$Step (b) Graph of $y = 0.5^{\left(\frac{x}{3}\right)} + 2$ for $-6 \leq x \leq 0$
Now, we are tasked with drawing the graph of the function $y = 0.5^{\left(\frac{x}{3}\right)} + 2$ for values of $x$ between $-6$ and $0$.
Plot the points from the completed table and join them smoothly to form the curve. The graph is a decreasing exponential curve approaching $y = 2$.
Step (c) Estimating the root of the equation
We are tasked with finding an estimate of the root of the equation
$\log_2\left((2x+2)^3\right) + x + 3 = 0$
This equation is equivalent to
$\log_2\left((2x+2)^3\right) = -x - 3$
$3\log_2(2x+2) = -x - 3$
$\log_2(2x+2) = -\frac{x}{3} - 1$
$2x+2 = 2^{-\frac{x}{3}-1}$
$2x+2 = 2^{-1\left(\frac{x}{3}+1\right)}$
$2x+2 = 0.5^{\frac{x}{3}+1}$
$2x+4 = 0.5^{\frac{x}{3}+1} + 2$
This equation simplifies to the form $y = 2x + 4$ and the corresponding function is $y = 0.5^{\frac{x}{3}+1} + 2$.
To find the root of the equation, we plot both functions $y = 2x + 4$ and $y = 0.5^{\frac{x}{3}+1} + 2$ on the same grid and determine their point of intersection.
From the graph, we can estimate the point of intersection of the two functions. Based on the graph, the approximate value of $x$ where both curves intersect is $x \approx -0.74$.
Thus, the root of the equation is approximately $x = -0.7$, correct to one decimal place.
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