Can someone tell me how to approach this, I could care less about the answer I just want to know how to solve this without Desmos. Thank you so much!

the vertex is at (4, -14)

Step-by-step explanation:

Review "quadratic equations."  These have three terms, and the highest power of x is x^2.  We need to identify the coefficients of the x^2, x and x^0 terms to get info needed to identify the axis of symmetry.

Given f(x) = x^2 - 8x + 2, we see that these coefficients are 1, -8 and 2.

                                                                            b

The formula for the axis of symmetry is x  = -

                                                                             2a

which here is x = -(-8)/[2*1], or x = 8/2  or x = 4:  the axis of symmetry.

To find the y-coordinate of the vertex, evaluate the function at x = 4:

f(4) = 4^2 - 8(4) + 2 = 16 - 32 + 2, or f(4) = -14.

Thus, the vertex is at (4, -14).  The graph is that of a parabola that opens up, and the parabola is symmetric about the vertical line x = 4.

72 is the answer to this problem