Supposed that f(x)= x^2+2x-8

Question

a) What are the coordinates of the vertex?

 

b) Find the x-intercepts using algebra.

 

Answer 1

f(x) = x² + 2x - 8

 

y= x² + 2x - 8

 

a=1,b=2,c=-8

 

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

 

subsitute the value of x in eq to find y coordiane

 

y= x² + 2x - 8

y=(-1)² +2(-1)-8

 

  =1-2-8=-9

 

so vertex is(-1,-9)

 

To find x-intercept

 

f(x) = x² + 2x - 8

 

x² + 2x - 8 =0

(x+4)(x-2)=0

x=-4 and x=2

x intercept is (-4,0) and (2,0)

Answer 2

f(x) = x² + 2x - 8

f(x) = x² + 2x + 1 - 1 - 8

f(x) = (x + 1)² - 9

 

x-intercept, f(x) = 0. You can do that.