VCAA Mathematical Methods Functions, relations and graphs

Incorrect. As $x$ approaches infinity, the value of $e^x$ grows without bound, meaning the limit is infinity, not $e$.

Incorrect. Any non-zero base raised to the power of 0 equals 1, so when $x = 0$, $y = e^0 = 1$, rather than $e$.

Incorrect. The function $y = e^x$ has a horizontal asymptote at $y = 0$ as $x$ approaches negative infinity, not a vertical asymptote at $x = 0$.

This incorrectly assumes the horizontal transformation is $2x + 3 = 1$, which would give $x = -1$. However, the argument is $2x - 3$.

This option uses an incorrect $x$-value by solving $2x + 3 = 1$ and an incorrect $y$-value calculation.

While this correctly identifies $x = 2$, it incorrectly calculates the $y$-value as $y = -1 + 3 = 2$ instead of $y = 1 - 3 = -2$.

This option is incorrect because while $f'(-1)=0$, the second derivative $f''(-1) = e^{-1}(-1+2) = e^{-1}$ is positive, not negative.

This option is incorrect because the first derivative evaluates to zero at $x=-1$, not a negative value, and the second derivative is positive.

This option is incorrect because the first derivative $f'(-1)$ equals $0$, not a value less than $0$.

This is the value of the function $N(3) = 15 \ln(22) \approx 46$. This represents the number of koalas (or the population increase) at year 3, rather than the rate at which the population is changing.

This value appears to be the result of calculating $N(3) - 20 \approx 26$. This subtracts the initial population from the model's value at $t=3$, which does not represent the instantaneous rate of change.

This is an incorrect value. It does not correspond to the derivative at $t=3$ or the function value, likely resulting from a calculation error.

The natural logarithm is undefined at $x = 0$ because there is no real power to which $e$ can be raised to yield 0.

The natural logarithm is not defined for negative numbers in the real number system.

This represents all real numbers, which is the domain of an exponential function like $y = e^x$, but the domain of $y = \ln(x)$ is restricted to strictly positive values.

This value corresponds to the vertical shift of the function ($-4$), which moves the graph up or down but does not affect the position of the vertical asymptote.

This would be the asymptote for the function $\log_9(x + 3)$. The expression $(x - 3)$ indicates a horizontal shift to the right, resulting in a positive value for the asymptote.

This option confuses the magnitude of the vertical shift with the asymptote location. The vertical asymptote is determined solely by the horizontal shift inside the logarithm's argument.

Dilation by $1/2$ from the $y$-axis gives $a^{2x}$, and translating left by $1$ unit gives $a^{2(x+1)} = a^{2x+2}$. This produces $g(x)$, so it is not the correct answer.

Dilation by $1/2$ from the $y$-axis gives $a^{2x}$, and dilating by $a^2$ from the $x$-axis gives $a^2 \cdot a^{2x} = a^{2x+2}$. This produces $g(x)$, so it is not the correct answer.

Dilating by $a^3$ from the $x$-axis gives $a^{x+3}$, translating right by $1$ unit gives $a^{(x-1)+3} = a^{x+2}$, and dilating by $1/2$ from the $y$-axis gives $a^{2x+2}$. This produces $g(x)$, so it is not the correct answer.

This incorrect value is half of the correct answer, which may result from a calculation error during the multiplication of fractions or evaluating the trigonometric ratio.

This answer results from evaluating $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ instead of $\cos\left(\frac{\pi}{3}\right)$ in the derivative, or incorrectly assuming the derivative of sine is sine.

This option is incorrect and likely results from misapplying the chain rule or arithmetic errors when combining the constants.

This result comes from incorrectly multiplying the given logarithm by 100 ($0.778 \times 100$). However, $\log_{10}(100 \times 6) = \log_{10} 6 + \log_{10} 100$, which involves addition, not multiplication.

This option incorrectly adds 10 to the given value. Since $600 = 6 \times 10^2$, you must add the exponent 2 (because $\log_{10} 100 = 2$), not the base 10.

This value represents $2 \times \log_{10} 6$, which equals $\log_{10}(6^2) = \log_{10} 36$. The logarithm of a product ($6 \times 100$) requires adding the logs, not multiplying the log value by 2.

This option only includes even integers, which misses half of the actual asymptotes that occur at every integer value.

This option only includes odd integers, missing all the even integer asymptotes that are also part of the graph.

This represents half-integer values, which would be the asymptotes if the function was not shifted by $\frac{1}{2}$ inside the argument.

This value is incorrect. It is close to $10^{-0.5}$ rather than the required $10^{-0.2}$.

This value is incorrect and does not result from solving the logarithmic equation for $[\text{H}^+]$.

This result comes from calculating $10^{0.2}$, which incorrectly ignores the negative sign in the pH formula.

The range of $\cos(9x)$ is $[-1, 1]$, so the range of $f(x) = 6 + 3\cos(9x)$ is $[6-3, 6+3] = [3, 9]$, which does not match the required range.

The range of $\cos(3x)$ is $[-1, 1]$, so the range of $f(x) = 6 + 6\cos(3x)$ is $[6-6, 6+6] = [0, 12]$, which does not match the required range.

The range of $\cos(3x)$ is $[-1, 1]$, so the range of $f(x) = 9 - 6\cos(3x)$ is $[9-6, 9+6] = [3, 15]$, which does not match the required range.