Question

Find the x- and y-intercepts of the parabola y=x^2+4x−6

Answer 1

The x-intercepts are when y=0, so:

`x^2 + 4x-6=0\\\\x=\frac{-4 \pm \sqrt{4^(2)-4(1)(-6)}}{2}\\\\x=(-4 \pm √(40))/(2)\\\\x=-2 \pm √(10)`

So, the x-intercepts are at
`\boxed{(-2 \pm √(10), 0)}`

The y-intercept is when x=0, so
`y=0^(2)+4(0)-6`, and thus the y-intercept is at (0, -6).