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

2 Answers
Mar 26, 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 = -1/4 - 2 = -9/4#
#Vertex (-1/2, -9/4)#.
To find y-intercept, make x = 0 --> y = -2
To find x-intercepts, make t = 0, and solve the quadratic equation:
#x^2 + x - 2 = 0.#
Since a + b + c = 0, use shortcut. The 2 x-intercepts (real roots) are:
1 and #c/a = -2#.

Mar 26, 2016

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

x int: #(-2,0)# and #(1,0)#

y int: #(0,-2)#

Explanation:

In a quadratic equation like #ax^2+bx+c# , the formula #-b/(2a)# gives us the #x# coordinate of the vertex. Then we put the #x# value we found in the equation to get the #y# value.

So, #x=-1/2# #-> f(x)=y=1/4-1/2-2 -> y=-9/4 -> v(-1/2, -9/4)#

To find the #x# intercepts, we should value #y# as #0# in the equation;

#0=x^2+x-2#
#0=(x+2)(x-1)#
#x=-2, x=1#

To find the #y# intercept, we should value #x# as #0# in the equation;

#y=-2#