Please help find vertex and intercepts of y=2((x-1)^2)-8

Please help find vertex and intercepts of y=2((x-1)^2)-8. Graph if possible, thanks

y = 2(x - 1)² - 8

The equation is in Vertex Form so the vertex is at (1, - 8).

x-int.: y = 0:

2(x - 1)² - 8 =0
2(x - 1)² = 8
(x - 1)² = 8 / 2
(x - 1)² =4
x - 1 = √4
x - 1 = ± 2
x = 1 ± 2

If x = 1 + 2,
x = 3

If x = 1 - 2,
x = - 1

x-int. (- 1, 0) and (3, 0)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯


y-int.: x = 0:

y = 2(0 - 1)² - 8
y = 2(- 1)² - 8
y = 2(1) - 8
y = 2 - 8
y = - 6

y-int. (0, - 6)
¯¯¯¯¯¯¯¯¯¯¯

The y intercept is the places where the line crosses the y axis
The x intercepts are the places the line crosses the x axis
here I think your equation: y=2(x(squared)-2x+1) - 8 = 2x(squared) -4x -6= (2x-6)(x+1)
Here you have parabola, it is upward facing.
When x = 3 and -1, then y=0, so x intercepts are (3,0) and( -1,0)
The vertex is (1,-8)

By intercepts I'm going to guess you mean the x- and y- axis intercepts, the vertex, however, is just (1, -8) as the general equation for vertex form is:

y = a(x - p)² + q where the vertex is (p, q)

the intercepts are always at (x, 0) and (0, y)

sub those values in to get your intercepts

(x, 0)

0 = 2(x - 1)² - 8
8 = 2(x - 1)²
4 = (x - 1)²
+-2 = x - 1
x = 3, -1

(3, 0)
(-1, 0)

(0, y)

y = 2(0 - 1)² - 8
y = 2(-1)² - 8
y = 2 - 8
y = -6

(0, -6)

Use www.wolframalpha.com for graphing!! It's really helpful :)

y=2((0-1)^2)-8
y=-6 is the y- intercept

make y=0 and solve for x to get x-intercepts soo...
0=2(x^2-2x+1)-8
0=2x^2-4x+2-8
0=2x^2-4x-6
0=(2x+2) (x-3)
So 2x+2 = 0 and x-3 = 0
x= -1, 3 (x-intercepts)

i dont know about the vertex...

y = 2(x - 1)^2 - 8 => already in the vertex form:
vertex at(1 , -8)
2(x - 1)^2 = 8
(x - 1)^2 = 4
x - 1 = ±2
x = -1 , 3 => x-intercepts at(-1 , 0) & (3 , 0)
y = 2(-1)^2 - 8 = -6 => intercept at (0 , -6)