Question

What are the vertex, axis of symmetry, maximum or minimum value, domain, and range of the function, and x and y intercepts for `y=x^2-10x+2`?

Answer 1

• `y=x^2-10x+2` is the equation of a parabola which will open upwards(because of the positive coefficient of `x^2`)
  So it will have a Minimum

• The Slope of this parabola is
  `(dy)/(dx) = 2x-10`
  and this slope is equal to zero at the vertex
  `2x - 10 = 0`
  `-> 2x = 10 -> x = 5`

• The X coordinate of the vertex will be `5`

`y=5^2-10(5)+2 = 25-50+2 =-23`
The vertex is at `color(blue)((5,-23)`

and has a Minimum Value `color(blue)(-23` at this point.

• The axis of symmetry is `color(blue)(x=5`

• The domain will be `color(blue)(inRR`(all real numbers)

• The range of this equation is `color(blue)({y in RR : y>=-23}`

• To get the x intercepts, we substitute y = 0
  `x^2-10x+2 = 0`
  We get two x intercepts as `color(blue)((5+sqrt23) and (5-sqrt23)`

• To get the Y intercepts, we substitute x = 0
  `y = 0^2 -10*0 + 2 = 2`
  We get the Y intercept as `color(blue)(2`

• This is how the Graph will look:
  graph{x^2-10x+2 [-52.03, 52.03, -26, 26]}