Question

what are the vertex and x-intercepts of the graph of y=x^2-6x-7

Answer 1

To find the vertex of the parabola represented by the equation y = x^2 - 6x - 7, we can first rewrite the equation in vertex form by completing the square.

y = (x^2 - 6x + 9) - 9 - 7
y = (x - 3)^2 - 16

In this form, the vertex of the parabola is at (3, -16).

To find the x-intercepts, we set y = 0 and solve for x:

0 = x^2 - 6x - 7
0 = (x - 7)(x + 1)

So, the x-intercepts are at x = 7 and x = -1.

Answer 2

alternate method (I think easier)
y = x^2 - 6x - 7 = (x-7)(x+1)
so y = 0 when x = 7 and x = -1
the vertex x value is halfway between those x axis intercepts so at x = (7-1)/2 = 3
when x = 3, y = (3-7)(3+1) = -4*4 = -16
so vertex at ( 3, -16)

Answer 3

Yes, your method is definitely more straightforward. Thank you for pointing that out.

To summarize:
- The x-intercepts are x = 7 and x = -1.
- The vertex is at (3, -16).