Question

Answer this please :(

Answer 1

Part A: check B, E and F.

Part B: check E and G.

Explanation:

The equation
`y = a(x - b)^2 + c` is the equation of a parabola written in the vertex form, where the vertex will be (b, c).

So, if the vertex is (2, -1), we have that b = 2 and c = -1

To find the c value, we use the information that the y-intercept is 3, so we have the point (0, 3). Using x = 0 and y = 3, we have:

`3 = a(0 - 2)^2 - 1`

`3 = 4a - 1`

`4a = 4`

`a = 1`

So we have a = 1, b = 2 and c = -1.

Part A: check B, E and F.

To find the x-intercepts, we need to find the values of x where y = 0:

`0 = (x - 2)^2 - 1`

`x^2 - 4x + 4 - 1 = 0`

`x^2 - 4x + 3 = 0`

Solving using Bhaskara's formula (a = 1, b = -4 and c = 3), we have:

`\Delta = b^2 - 4ac = 16 - 12 = 4`

`x_1 = (-b + √(\Delta))/2a = (4 + 2)/2 = 3`

`x_2 = (-b - √(\Delta))/2a = (4 - 2)/2 = 1`

So the x-intercepts are 1 and 3

Part B: check E and G.