Question

Explain the procedure (but don’t create a graph) for graphing the quadratic function in vertex form f(x)=2(x-3)^2-1 using a pencil and paper. Your explanation should include an understanding of the perfect square patterns from the vertex. Explain the entire proccess

Answer 1

The procedure in the explanation

Explanation:

we have

`f(x)=2(x-3)^2-1`

This is the equation of a vertical parabola written in vertex form open upward (the leading coefficient is positive)

The vertex is a minimum

The vertex is the point (3,-1)

To graph the parabola find the axis of symmetry and the intercepts

Find the axis of symmetry

The axis of symmetry of the vertical parabola is equal to the x-coordinate of the vertex

so

`x=3`

Find the y-intercept

The y-intercept is the value of y when the value of x is equal to zero

so

For x=0

`f(x)=2(0-3)^2-1`

`f(x)=17`

The y-intercept is the point (0,17)

Find the x-intercepts

The x-intercept is the value of x when the value of y is equal to zero

so

For y=0

`0=2(x-3)^2-1`

solve for x

`(x-3)^2=(1)/(2)`

square root both sides

`x-3=\pm(1)/(√(2))`

`x=3\pm(1)/(√(2))`

`x=3\pm(√(2))/(2)`

The x-intercepts are the points

`(3-(√(2))/(2),0)\ and\ (3+(√(2))/(2),0)`

`(2.293,0)\ and\ (3.707,0)`

To graph the quadratic equation, plot the vertex, the axis of symmetry, the y-intercept and the x-intercepts and join them

see the attached figure to better understand the problem