Question

How do you find the vertex and the intercepts for `y=x^2-10x+20`?

Answer 1

Vertex is at `(5,-5)` y-intercept is at `(0,20)` x-intercepts are at `(2.764,0) and (7.236,0)`

Explanation:

`y=x^2-10x+20 = (x-5)^2 -25+20 =(x-5)^2-5 :.`Comparing with vertex form of equation; `y=a(x-h)^2+k` we get vertex at `(h,k)=(5,-5)`.To get y-intercept putting x=0 in the equation we get `y=20` and to find x-intercepts putting y=0 in the equation we get `(x-5)^2-5=0 or (x-5)^2 =5 or (x-5) = +-sqrt5 or x= 5+-sqrt5 or x= 2.764,7.236 :.`Vertex is at `(5,-5)` y-intercept is at `(0,20)` x-intercepts are at `(2.764,0) and (7.236,0)` graph{x^2-10x+20 [-40, 40, -20, 20]}[Ans]