Question

How do you graph `y=2(x+1)^2+3`?

Answer 1

See the explanation.

Explanation:

By 'how do you graph' I am assuming the question is really asking: "how do you find the critical points enabling the sketching of the equation...?"

The given equation is the consequence of 'completing the square' and is often referred to as 'vertex form equation'

`color(blue)("Determine the vertex")`

Given: `y=2(x color(red)(+1))^2color(green)(+3)`

Where
`x_("vertex")=(-1)xxcolor(red)(+1)=-1`
`y_("vertex")=color(green)(+3)`

`=>` Vertex`->(x,y)=(-1,3)`
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`color(blue)("Determine the y-intercept")`

Set `x=0` because the y-axis crosses the x-axis at `x=0`

`y=2(0+1)^2+3" "=" "5`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the x-intercept")`

Set `y=0` because the x-axis crosses the y-axis at `y=0`

`0=2(x+1)^2+3`

Subtract 3 from both sides

`-3=2(x+1)^2`

Divide both sides by 2

`-3/2=(x+1)^2`

Square root both sides

`sqrt(-3/2)=x+1`

As we are square rooting a negative value it means that the curve does NOT cross or touch the x-axis

This is logical because:

As the coefficient of `x^2` is positive the graph is of form `uu`
The vertex is at `y=3` which is above the x-axis.
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Any other points you wish to determine involves building a table of values. Marking the related points and then drawing the best fit you can.