Question

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

Answer 1

the `x`-intercepts are `-5` and `-3`, the y-intercept is `15` and the vertex is `(-4,-1)`

Explanation:

Because the equation is in the form `y=a(x-p)(x-q)`, we know the `x`-intercepts are `p` and `q`
Therefore, the `x`-intercepts are at `-5` and `-3`
Note the negative signs and the fact that in the case `a=1` so it is committed.
This makes sense because the `x`-intercepts are whenever `y=0`. This occurs when either bracket is `0`.

The `x`-value of the vertex is the midvalue of the intercepts.
let `V_x` be the `x`-value of the vertex
`V_x=((-5)+(-3))/2`
`V_x=-4`
Subbing in `x=-4` to solve for the `y`-value of the vertex:
Let this be `V_y`:
`V_y=((-4)+5)((-4)+3)`
`V_y =(1)(-1)`
`V_y = -1`

Therefore, the vertex is `(-4,-1)`

To find the `y`-intercept we substitute `x=0` into the equation.
`y=(5)(3)`
`y=15`

In conclusion, the `x`-intercepts are `-5` and `-3`, the `y`-intercept is `15` and the vertex is `(-4,-1)`

Included below are graphs of the equation:
graph{(x+5)(x+3) [-7.757, 3.6, -2.454, 3.224]}
graph{(x+5)(x+3) [-49.16, 45, -23.16, 23.9]}