Question

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

Answer 1

`color(red)("Vertex at " (-2, -12))`
`color(red)("The intercepts are " (0, 0) and (-4, 0)`

Explanation:

From the given equation
`y=3x^2+12x`

We transform the equation by Completing the square method

`y=3x^2+12x`
factor out 3

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

The coefficient 4 will be divided by 2 and the result will be squared giving 4. This is the number to be added and subtracted inside the grouping symbol.

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

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

Now notice the `x^2+4x+4` is a perfect square trinomial reducible to `(x+2)^2`, so that

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

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

Distribute the 3 back

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

transpose the -12 to the left of the equation

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

divide both sides by 3

`1/3(y+12)=(x+2)^2`

We now have the Vertex Form

`(x--2)^2=1/3(y--12)`

with `color(red)("Vertex at " (-2, -12))`

To solve for the intercepts, we will use the given general form
`y=3x^2+12x`

Intercepts are points of intersection of the curve with the axes.

To solve for the x-intercept, we set `y=0` then find `x` values
`y=3x^2+12x`
`0=3x^2+12x`
We can solve by factoring
`0=3x(x+4)`
We have two factors 3x and x+4 to be equated to 0.

`3x=0`

`x=0/3`
`x=0`

the other one

`x+4=0`
`x=-4`

The x-intercepts are `(0, 0)` and `(-4, 0)`

Let us now solve the y-intercept by setting `x=0` this time, and solving for `y` values

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

Therefore, the y-intercept is the same point `(0, 0)`.

`color(red)("The intercepts are " (0, 0) and (-4, 0)`
`color(red)("Vertex at " (-2, -12))`

Kindly inspect the graph of the equation and check the points that we solved

graph{(x--2)^2=1/3(y--12)[-30,30,-15,15]}

God bless....I hope the explanation is useful.