Given
The function is defined as:
[tex]y\text{ = 2\lparen x -3\rparen}^2\text{ - 1}[/tex]x-intercepts
The x-intercepts of the function y are the values of x when y = 0
Substituting 0 for y and solving for x
[tex]\begin{gathered} 2(x-3)^2\text{ -1 = 0} \\ 2(x-3)^2\text{ = 1} \\ Divide\text{ both sides by 2} \\ (x-3)^2\text{ = }\frac{1}{2} \\ Square\text{ root both sides} \\ x-3\text{ = }\pm\sqrt{\frac{1}{2}} \\ x\text{ = 3 }\pm\text{ }\sqrt{\frac{1}{2}} \end{gathered}[/tex]Hence, the x-intercepts are:
[tex](\sqrt{\frac{1}{2}}\text{ + 3, 0\rparen, \lparen-}\sqrt{\frac{1}{2}}\text{ + 3,0\rparen}[/tex]y-intercepts
The y-intercepts are the values of y when x = 0
[tex]\begin{gathered} y\text{ = 2\lparen0-3\rparen}^2-\text{ 1} \\ =\text{ 2}\times9-1 \\ =\text{ 17} \end{gathered}[/tex]Hence, the y-intercept is (0, 17)
Maximum or minimum of the function
The given equation is in vertex form.
[tex]\begin{gathered} y\text{ = a\lparen x-h\rparen}^2\text{ + k} \\ Where\text{ \lparen h,k\rparen is the vertex} \end{gathered}[/tex]Hence, the minimum value of the function is (3,-1)