Consider the quadratic function y2x25x 1 a Find the coordin

Consider the quadratic function y=2x2-5x + 1 (a) Find the coordinates of the vertex. (2 points) (b) Find the x-intercepts, if any. (2 points) (c) Find the range of the function. (2 points) (d) Graph the function. (4 points)

Solution

a. y =2x²-5x+1

Vertex,x =-b/2a= -(-5)/(2(2)= 5/4

y=2(5/4)²-5(5/4)+1 = (50/16)-(25/4)+1= (50-100+16)/16

=-34/16= -17/8

(5/4,-17/8)

b. x intercepts

2x²-5x+1=0

2x²-4x-x+1=0

2x(x-2)-1(x-2)=0

x=1/2, 2

(1/2,0),(2,0)

c. Y values cant be less than -17/8

Therefore range [-17/8,infinity)