Question

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

Answer 1

See explanation.

Explanation:

Finding the vertex:
Use the formula for the point of the vertex `(-b/(2a),f(-b/(2a)))`.

Remember that the quadratic is in the general form `ax^2+bx+c`, so `a=1` and `b=2`.

`-b/(2a)=-1`

`f(-b/(2a))=f(-1)=(-1)^2+2(-1)-5=-6`

The vertex is at the point `(-1,-6)`.

Finding the y-intercept:

Plug in `0` for `x`.

`y=(0)^2+2(0)-5=-5`

The `y`-intercept is `(0,-5)`.

Finding the x-intercept:

Plug in `0` for `y`.

`0=x^2+2x-5`

`x=(-2+-sqrt(2^2-4(1)(-5)))/(2(1))`

`x=(-2+-sqrt24)/2`

`x=(-2+-2sqrt6)/2=-1+-sqrt6`

The `x`-intercepts are `(-1+sqrt6,0)` and `(-1-sqrt6,0)`.