Answer:
The procedure in the explanation
Step-by-step explanation:
we have
[tex]f(x)=2(x-3)^2-1[/tex]
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
[tex]x=3[/tex]
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
[tex]f(x)=2(0-3)^2-1[/tex]
[tex]f(x)=17[/tex]
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
[tex]0=2(x-3)^2-1[/tex]
solve for x
[tex](x-3)^2=\frac{1}{2}[/tex]
square root both sides
[tex]x-3=\pm\frac{1}{\sqrt{2}}[/tex]
[tex]x=3\pm\frac{1}{\sqrt{2}}[/tex]
[tex]x=3\pm\frac{\sqrt{2}}{2}[/tex]
The x-intercepts are the points
[tex](3-\frac{\sqrt{2}}{2},0)\ and\ (3+\frac{\sqrt{2}}{2},0)[/tex]
[tex](2.293,0)\ and\ (3.707,0)[/tex]
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
