How do you find the vertex and intercepts for #y = x^2 +2x - 5#?

1 Answer
Nov 24, 2015

See explanation.

Explanation:

Finding the vertex:
Use the formula for the point of the vertex #(-b/(2a),f(-b/(2a)))#.

Remember that the quadratic is in the general form #ax^2+bx+c#, so #a=1# and #b=2#.

#-b/(2a)=-1#

#f(-b/(2a))=f(-1)=(-1)^2+2(-1)-5=-6#

The vertex is at the point #(-1,-6)#.

Finding the y-intercept:

Plug in #0# for #x#.

#y=(0)^2+2(0)-5=-5#

The #y#-intercept is #(0,-5)#.

Finding the x-intercept:

Plug in #0# for #y#.

#0=x^2+2x-5#

#x=(-2+-sqrt(2^2-4(1)(-5)))/(2(1))#

#x=(-2+-sqrt24)/2#

#x=(-2+-2sqrt6)/2=-1+-sqrt6#

The #x#-intercepts are #(-1+sqrt6,0)# and #(-1-sqrt6,0)#.