Question

What are the important points needed to graph `y = 3x^2 + 8x - 6`?

Answer 1

Its vertex is `((-4)/3, (-2)/3)`
Since the co-efficient of `x^2` is positive, the curve is open upwards.

It has a minimum at `((-4)/3, (-2)/3)`
Its y- intercept is `-6`

Explanation:

Given-

`y=3x^2+8x-6`

We have to find the vertex

`x=(-b)/(2a)=(-8)/(2 xx 3)=(-8)/6=(-4)/3`

At `x=(-4)/3`;

`y=3((-4)/3)^2+8((-4)/3)-6`
`y=3((16)/9)-32/3-6`
`y=48/3-32/3-6=(-2)/3`

Its vertex is `((-4)/3, (-2)/3)`

Take two points on either side of `x=(-4)/3`

Find the y values. Plot the points. Join them with a smooth curve.

Since the co-efficient of `x^2` is positive, the curve is open upwards.

It has a minimum at `((-4)/3, (-2)/3)`

Its y- intercept is `-6`

Since the co-efficient of `x^2` is 3, the curve is a narrow one.

graph{3x^2+8x-6 [-25.65, 25.65, -12.83, 12.82]}