Question

How do you find the vertex and the intercepts for `y = (x + 5)^2 +2`?

Answer 1

vertex: `(-5, 2)`
no `x`-intercepts
`y`-intercept: `(0, 27)`

Explanation:

Standard form used for graphing a parabola: `y = a(x-h)^2+k`,
where vertex:`(h, k)`

Finding the vertex:
Since the function `y = (x+5)^2 + 2` is in standard form, you can easily find the vertex to be `(-5, 2)`.

Finding `x-`intercepts by letting `y = 0`:
1. Distribute the function: `y = x^2+5x + 5x +25+2`
2. Add like terms: `y = x^2 + 10x + 27`
3. Factor or use the quadratic formula `x =( -B+- sqrt(B^2-4AC))/(2A)`
`x = (-10+- sqrt(100-4*1*27))/2 = (-10+-sqrt(-8))/2 = -5 +-sqrt(2)i`

No real solutions, only complex solutions. This means the function does not touch the `x-`axis. No `x-`intercepts

Finding `y`-intercepts: by letting `x = 0`:
`y = 5^2 + 2 = 27`
`y`-intercept: `(0, 27)`