Question

How do you find the vertex and intercepts for `y = 3x^2 + 7x – 4`?

Answer 1

The intercepts are given by attributing zero to one of the variables, while the vertex has its coordinates `x_v` and `y_v` given by formulae.

Explanation:

The intercepts:

1. When `x=0`, `y=3(0)^2+7(0)-4`, then when `x=0`, `y=-4`

Thus, the first intercept, which corsses the axle `y` has the following coordinates: `(0,-4)`

1. When `y=0`, we must resort to Bhaskara, which will go like this:

`(-7+-sqrt((7)^2-4(3)(-4)))/(2(3)) = (-7+-(sqrt(97)))/6`

Let's approximate `sqrt(97)~=9.84`

`(-7+-9.84)/6 => x_1~=0.473` and `x_2~=-2.807`

Thus, the intercepts that cross the axis `x` (considering an approximation) are `(0.473,0)` and `(-2.807,0)`

The vertex has the following formulae:

`x_v=-b/(2a)=-7/6~=-1.167`

`y_v=-Delta/(4a)=-97/12~=-8.083`

Vertex's coordinates: `(-1.167,-8.083)`