Question

Someone please help me I need this for tomorrow I don’t get it at all PLEASE

Answer 1

Answer and Explanation:

We are being asked to graph parabolas given in vertex form equations:

• 
  `y=m(x-a)^2 + b`

where
`(a,b)` is the parabola's vertex and
`m` is its vertical stretch from a standard parabola (when
`m = 1`).

Also note that if m is negative, then the parabola opens downward.

A standard parabola (
`y=x^2`) has the values:

`\begin{array} c \cline{1-2}x & y \\ \cline{1-2} -3 & 9 \\ \cline{1-2} -2 & 4\\ \cline{1-2} -1 & 1\\ \cline{1-2} 0 & 0 \\ \cline{1-2} 1 & 1\\ \cline{1-2} 2 & 4 \\ \cline{1-2} 3 & 9 \\ \cline{1-2} \end{array}`

which corresponds to the attached graph.

For a stretched parabola (
`y=2x^2`), the values look like:

`\begin{array} \cline{1-2}x & y \\ \cline{1-2} -3 & 18 \\ \cline{1-2} -2 & 8\\ \cline{1-2} -1 & 2\\ \cline{1-2} 0 & 0 \\ \cline{1-2} 1 & 2\\ \cline{1-2} 2 & 8 \\ \cline{1-2} 3 & 18 \\ \cline{1-2} \end{array}`

For a shifted parabola (
`y=-(x+3)^2+2`), the values look like:

`\begin{array} \cline{1-2}x & y \\\cline{1-2} -6 & -7\\\cline{1-2} -5 & -2\\\cline{1-2} -4 & 1\\\cline{1-2} -3 & 2 \\\cline{1-2} -2 & 1\\\cline{1-2} -1 & -2 \\\cline{1-2} 0 & -7 \\ \cline{1-2} \end{array}`