Question

How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of `y=x^2-10x+2`?

Answer 1

Vertex is `(5,-23)`
Axis of symmetry `x=5`
Y-intercept `y=2`
X- intercepts are `(9.8, 0) and (0.2, 0)`

Explanation:

Vertex

`3x=(-b)/(2a)=(-(-10))/(2 xx 1)=10/2=5`

At `x=5`

`y=5^2-10(5)+2`
`y=25-50+2`
`=27-50`
`y=-23`
Vertex is `(5,-23)`
Axis of symmetry `x=5`

Y-intercept

At `x=0`

`y=0^2-10(0)+2`
`y=2`

`b^2-(4ac)= (-10^2)-(4xx1xx2)`

X-intercept

At `y=0`

`x^2-10x+2=0`

`x=(-b+- sqrt(b^2-(4ac)))/(2a)`
`x=(-(-10)+- sqrt((-10)^2-(4 xx 1 xx 2)))/(2 xx 1)`
`x=(10+- sqrt(100-8))/(2)`
`x=(10+- sqrt(92))/(2)`
`x=(10+- 9.6)/(2)`
`x=(10+9.6)/2=19.6/2=9.8`
`x=(10-9.6)/2=0.4/2=0.2`

X- intercepts are `(9.8, 0) and (0.2, 0)`

graph{x^2-10x+2 [-58.5, 58.55, -29.25, 29.25]}