How do you find the vertex and the intercepts for #f(x)=x^2+x-2#?

1 Answer
Mar 14, 2016

#vertex (-1/2, -9/4)#

Explanation:

x-coordinate of vertex:
#x = -b/(2a) = - 1/2#
y-coordinate of vertex:
#y(-1/2) = 1/4 - 1/2 - 2 = - 9/4#
#Vertex (-1/2, -9/4)#.
To find y-intercept, make x = 0 --> y = -2.
To find x-intercepts, make y = 0 and solve the quadratic equation
#y = x^2 + x - 2 = 0#
Since a + b + c = 0, use shortcut.
The 2 real roots are 1 and #c/a = -2.#