Consider the Quadratic Function f(x)=2x^2-4x+1

Question

Write the quadratic function in standard form by completing the square. Show how you arrived at your answer

(b) What is the vertex?

(c) What is the axis of symmetry? Be sure to write you answer as an equation.

(d) What is the y-intercept? Write your answer as a point.



(e) Find the x-intercepts algebraically. Show how you arrived at your answer. Write your answer as points.

Answer 1

2x^2 -4x + 1= 0

maximum point is when the first derivative = 0

first derivative is 4x -4 = 0

4x=4

x = 1

and y = 2(1)^2 -4(1) +1 = 2-4+1 =-1

vertex is (1,-1)

2x^2 -4x +1

divide by 2 and set = 0

x^2 -2x + 1/2 = 0

add 1/2 of the x term's coefficent to both sides , squared (1/2 x 2 = 1)^2 = 1

x^2 -2x + 1 + 1/2 = 0+1

factor

(x-1)^2 + 1/2 = 1

(x-1)^2 = = 1 -1/2 = 1/2

take the square root of both sides

x-1 = sqr(1/2)

add 1 to both sides

x = 1 + or - sqr(1/2)

x = 1 + or - sqr(1/2) = 1 + or - sqr1/sqr2 = 1 + or - sqr2/2

= 1 + or - (1.414/2)

= 1 + or - (0.707) = 1.707 or 0.293

axis of symmetry is x = 1, the real x value without the + or - sqr number

y-intercept is when x=0

x^2 -4x +1 = y

1= y = y-intercept the point is (0,1)

the x-intercepts are when y=0

0=x^2 -4x +1

That's about 1.7 and 2.3 solved above initially by completing the square