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

1 Answer

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)