Question

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Answer 1

lemme test say the 1st point \(\bf h(x)=y =3x^2\qquad \begin{array}{lllll}
&x&y\\
&(1\quad ,&3)
\end{array}\implies y = 3(1)^2\implies y = 3\) so what do you think, does that point lie on the function?

Answer 2

when x = 1, "y" end up being 3, so, we got the point ( 1, 3) so

Answer 3

lemme test the 2nd point \(\bf h(x)=y =3x^2\qquad \begin{array}{lllll}
&x&y\\
&(1\quad ,&9)
\end{array}\implies y = 3(1)^2\implies y = 3\) so we end up with the coordinate values of ( 1, 3) again... we did not get ( 1, 9) though

so what do you think, does ( 1, 9) lie on the function?

Answer 4

you see, you firstly use the value of "x" in the function, see what "y" gets in the end, if it matches up the pair of coordinates given, then it's on the function

Answer 5

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIzeF4yIiwiY29sb3IiOiIjRjAxNjE2In0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTQuOTQ0IiwiNS40NTYiLCItMC4zMzU5OTk5OTk5OTk5OTk0IiwiNi4wNjM5OTk5OTk5OTk5OTkiXX1d <--- notice the graph, those points that make up the line, are the ones that ARE on the function

so when you plug a value given for "x", the "y" value is whatever \(3x^2\) will spit out, and those 2 coordinates are indeed on the function