Question

Hi, can you help me to solve exercise #3 please !!

Answer 1

The equation of a parabola with vertex (h,k) is:

`f(x)=a(x-h)^2+k`

Where a is a constant.

Rewrite the given expression in vertex form by completing the square:

`\begin{gathered} f(x)=2x^2-2x+1 \\ =2(x^2-x)+1 \\ =2(x^2-x+(1)/(4)-(1)/(4))+1 \\ =2(x^2-x+(1)/(4))-2\cdot(1)/(4)+1 \\ =2(x-(1)/(2))^2-(1)/(2)+1 \\ =2(x-(1)/(2))^2+(1)/(2) \end{gathered}`

Use the expression in vertex form to answer parts a to c.

a)

To find the y-intercept, evaluate f at x=0:

`f(0)=2(0)^2-2(0)+1=1`

The axis of symmetry is a vertical line that passes through the vertex. Since the coordinates of the vertex are (1/2,1/2), then the equation of the axis of symmetry is:

`x=(1)/(2)`

And the x-coordinate of the vertex is 1/2.

b)

Use the values x=0, x=0.5, x=1, x=1.5, x=2 to make a table: