Question

How do you find the vertex and intercepts for `y = 3x^2 + 12x + 5`?

Answer 1

Vertex

The vertex is easily determined by completing the square.

`y = 3(x^2 + 4x + p) + 15`

`p = (b/2)^2`

`p = (4/2)^2`

`p = 4`

`y = 3(x^2 + 4x + 4 - 4) + 15`

`y = 3(x^2 + 4x + 4) - 4(3) + 15`

`y = 3(x + 2)^2 - 12 + 15`

`y = 3(x + 2)^2 + 3`

In vertex form, `y = a(x - p)^2 + q`, the vertex is located at the point `(p, q)`

Therefore, the vertex is at `(-2, 3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Intercepts:

y intercept:

`y = 3(0)^2 + 12(0) + 5`

#y = 5 -> (0, 5)

x intercept:

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

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

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

`O/ = x`

Therefore, there is no x intercept.

The graphical representation of this function proves that there is no x intercept.

graph{y = 3(x + 2)^2 + 3 [-22, 21.99, -11, 11]}

We could have also found, by intuition, that there are no x intercepts: In a quadratic function `y = a(x - p)^2 + q`, the parameter a represents the breadth and the direction of opening of the parabola. If `a > 0`, then the parabola opens up. This is our case. Since the vertex was at `(-2, 3)`, which is located in the second quadrant (above the x axis), and the parabola opens up, the function will never cross the x axis.

I hope you have learned a lot and that my answer was of great help.