Question

How do you find the axis of symmetry, vertex and x intercepts for `y=-x^2-3x+4`?

Answer 1

axis of symmetry is `x = -3/2`
vertex minimum at `25/4`
`x`-intercepts are `(1,0) and (-4,0)`

Explanation:

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

to find axis of symmetry and vertex, we can use completing a square to solve it.

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

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

`y = - (x + 3/2)^2 + 9/4 + 16/4`

`y = - (x + 3/2)^2 + 25/4` `->i`

it axis of symmetry can be found when `(x + 3/2) = 0`, then `x = -3/2`

since `- (x + 3/2)^2` is always has a -ve value, therefore it has a maximum vertex at `25/4`

to find `x`-intercepts, plug in `y = 0` in `-> i`

`0 = - (x + 3/2)^2 + 25/4`

`(x + 3/2)^2 = 25/4`

`(x + 3/2) = +- sqrt (25/4)`

`x = - 3/2 +- 5/2`

`x_1 = - 3/2 + 5/2 = 2/2 = 1`

`x_2 = - 3/2 - 5/2 = -8/2 = -4`

therefore `x`-intercepts are `(1,0) and (-4,0)`