Question

How do you find the vertex and the intercepts for `y= -3(x-1)(x+5)`?

Answer 1

Vertex is at `(-2, 27)`
Intercepts at `(1, 0)`, `(-5, 0)`,`(0, 15)`

Explanation:

From the given `y=-3(x-1)(x+5)`
Expand the right side so that it will take the form `y=ax^2+bx+c`

`y=-3(x-1)(x+5)`

`y=-3(x^2+4x-5)`

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

Now it takes the form `y=ax^2+bx+c`

with `a=-3` and `b=-12` and `c=15`

We can now solve for the vertex `(h, k)` using the formula

`h=-b/(2a)` and `k=c-b^2/(4a)`

`h=(-(-12))/(2(-3))=12/-6=-2`

`k=c-b^2/(4a)=15-(-12)^2/(4(-3))=15+144/12=15+12=27`

Vertex `(h, k)=(-2, 27)`

To solve for the intercept, let us use the given `y=-3(x-1)(x+5)`

Solve for x-intercept by setting `y=0`

`y=-3(x-1)(x+5)`

`0=-3(x-1)(x+5)`

`0=(x-1)(x+5)`

Equate both factors to 0
First factor
`(x-1)=0`
`x=1` there is an intercept at `(1, 0)`

Second factor
`x+5=0`
`x=-5` there is an intercept at `(-5, 0)`

Use again the original equation `y=-3(x-1)(x+5)` to solve for y-intercept and setting `x=0`

`y=-3(x-1)(x+5)`

`y=-3(0-1)(0+5)`

`y=-3(-1)(5)`

`y=15`

there is an intercept at `(0, 15)`

The graph of `y=-3(x-1)(x+5)` where you can see the
vertex `-2, 27)`, and intercepts at `(1, 0)`, `(-5, 0)`,`(0, 15)`

graph{y=-3(x-1)(x+5)[-60,60,-30,30]}

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