Question $1$ : Which functions have $(12,3)$ as a solution? (Select ALL that apply): $\newline$ (Select all that apply.) $\newline$ $y=4 x$ $\newline$ $y=0.25 x$ $\newline$ $y=x+9$ $\newline$ $y=x^{4}$ $\newline$ $y=x-9$ $\newline$ $y=\frac{1}{4} x$ $\newline$ $y=(x \div 6)+1$ $\newline$ $y=x \div 4$ $\newline$ Select all the functions whose graphs include the point $(12,3)$ . $\newline$ You may the whiteboard to work out any of your solutions / show your reasoning.

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Q. Question

1

: Which functions have

(12,3)

as a solution? (Select ALL that apply):

\newline

(Select all that apply.)

\newline

y=4 x

\newline

y=0.25 x

\newline

y=x+9

\newline

y=x^{4}

\newline

y=x-9

\newline

y=\frac{1}{4} x

\newline

y=(x \div 6)+1

\newline

y=x \div 4

\newline

Select all the functions whose graphs include the point

(12,3)

\newline

You may the whiteboard to work out any of your solutions / show your reasoning.

Question Prompt: Question Prompt: Identify which functions have $(12,3)$ as a solution.
Check $y=4x$ : Check $y=4x$ : $\newline$ Substitute $x=12$ into $y=4x$ : $\newline$ $y = 4 \times 12 = 48$ . $\newline$ Since $48 \neq 3$ , $(12,3)$ is not a solution for $y=4x$ .
Check $y=0.25x$ : Check $y=0.25x$ : $\newline$ Substitute $x=12$ into $y=0.25x$ : $\newline$ $y = 0.25 \times 12 = 3$ . $\newline$ Since $3 = 3$ , $(12,3)$ is a solution for $y=0.25x$ .
Check $y=x+9$ : Check $y=x+9$ : $\newline$ Substitute $x=12$ into $y=x+9$ : $\newline$ $y = 12 + 9 = 21$ . $\newline$ Since $21 \neq 3$ , $(12,3)$ is not a solution for $y=x+9$ .
Check $y=x^{4}$ : Check $y=x^{4}$ : $\newline$ Substitute $x=12$ into $y=x^{4}$ : $\newline$ $y = 12^4 = 20736$ . $\newline$ Since $20736 \neq 3$ , $(12,3)$ is not a solution for $y=x^{4}$ .
Check $y=x-9$ : Check $y=x-9$ : $\newline$ Substitute $x=12$ into $y=x-9$ : $\newline$ $y = 12 - 9 = 3$ . $\newline$ Since $3 = 3$ , $(12,3)$ is a solution for $y=x-9$ .
Check $y=\frac{1}{4}x$ : Check $y=\frac{1}{4}x$ : $\newline$ Substitute $x=12$ into $y=\frac{1}{4}x$ : $\newline$ $y = \frac{1}{4} \times 12 = 3$ . $\newline$ Since $3 = 3$ , $(12,3)$ is a solution for $y=\frac{1}{4}x$ .
Check $y=\frac{x}{6}+1$ : Check $y=\frac{x}{6}+1$ : $\newline$ Substitute $x=12$ into $y=\frac{x}{6}+1$ : $\newline$ $y = \left(\frac{12}{6}\right) + 1 = 2 + 1 = 3$ . $\newline$ Since $3 = 3$ , $(12,3)$ is a solution for $y=\frac{x}{6}+1$ .
Check $y=\frac{x}{4}$ : Check $y=\frac{x}{4}$ : $\newline$ Substitute $x=12$ into $y=\frac{x}{4}$ : $\newline$ $y = \frac{12}{4} = 3$ . $\newline$ Since $3 = 3$ , $(12,3)$ is a solution for $y=\frac{x}{4}$ .