Question

How do you use the important points to sketch the graph of `y=3x^2+6x-1`?

Answer 1

y-intercept`= -1`
`x_("intercept") =-1+-sqrt(48)/6` Exact value

`x_("intercept")~~0.155` to 3 decimal places approx. value
`x_("intercept")~~-2.155` to 3 decimal places approx. value

Vertex`->(x,y)=(-1,-4)`

Explanation:

Forgive the wording but this is a common mistake. Sketch does not mean all prettied up diagram. The actual diagram should take about 30 to 45 seconds.

When I used to do this I did not even use a ruler.

Consider the standardised form `y=ax^2+bx+c`
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`color(blue)("y-intercept" = c = -1)`
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There are various options for determining vertex and x-intercepts.

`color(brown)("Option 1:")`
Determine the x intercepts and `1/2` between is `x_("vertex")`

`color(brown)("Option 2:")`

Complete the square and with a small adjustment read of the coordinates of the vertex.

Progress the resulting equation to determine the x-intercepts

`color(brown)("Option 3:")`

Factorise the equation and thus determine x-intercepts.
`x_("vertex")" is " 1/2` way between.
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Consider the given `y=3x^2+6x-1`

Factorisation is not easy so lets revert to other methods
If the question proposer is UK then you need to memorise both completing the square and the traditional formula method.

Given that `y=ax^2+bx+c` then `x=(-b+-sqrt(b^2-4ac))/(2a)`

where `a=3; b= 6 ; c=-1`

`x=(-6+-sqrt(6^2-4(3)(-1)))/(2(3))`

`x=(-6+-sqrt(36+12))/6`

`x_("intercept") =-1+-sqrt(48)/6` Exact value

`x_("intercept")~~0.155` to 3 decimal places approx. value
`x_("intercept")~~-2.155` to 3 decimal places approx. value

We could find the mid point for x-vertex but let me show you a cool trick:

Write as: `y=3(x^2 color(red)(+6/3) x)-1`

`x_("vertex")=(-1/2)xx(color(red)(+6/3)) = - 1`

Substitute in original equation.

`y_("vertex")=3(-1)^2+6(-1)-1= 3-6-1= -4`

Vertex`->(x,y)=(-1,-4)`