Question

How do you find the x and y intercepts for `y = -2(x-1)^2 + 3`?

Answer 1

`y` intercept is 1
`x` intercepts are 2.22 and -0.22

Explanation:

The `y` intercept is when `x =0`
`=>` `y = -2(0 -1)^2 +3`

`=>` `y = -2` `xx` 1 +3#

`=>` `y = -2 +3`

`=>` `y = 1`

The `x` intercept is when `y = 0`
`=>` `0 = -2(x -1)^2 +3`

`=>` `0 = -2 (x^2 -2x +1) +3`

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

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

Use the formula `x=(-b\pm sqrt(b^2 -4ac))/(2a)`

`x=(4\pm sqrt(16+8))/(4)`

`x=(4 +sqrt(24))/4` or `x = (4 - sqrt(24))/4`
`x = 2.22474487` or `x = -0.22474487`

Answer 2

`p_(x_1)(1+sqrt(3/2)|0) or p_(x_2)(1-sqrt(3/2)|0)`
`p_y(0|1)`

Explanation:

Lets start with the interception of the y-axis. At the point where the graph intercepts this axis, `x=0`. Furthermore, it is the only point where `x=0`.

`y = -2(x-1)^2 + 3`
`x=0`
`y = -2(0-1)^2 + 3`
`y = -2 + 3=1`
`p_y(0|1)`

It works the same way for the x-axis. The only difference is that `y=0` instead of `x=0`.

`0 = -2(x-1)^2 + 3|-3`
`-3 = -2(x-1)^2 |:(-2)`
`3/2=(x-1)^2|sqrt()`
`+-sqrt(3/2)=x-1|+1`
`1+-sqrt(3/2)=x`
`p_1(1+sqrt(3/2)|0) or p_2(1-sqrt(3/2)|0)`

graph{y=-2(x-1)^2+3 [-2, 4, -5, 5]}