Question

What is the axis of symmetry and vertex for the graph `y=x^2-10x+2`?

Answer 1

vertex = (5,-23) , x = 5

Explanation:

The standard form of a quadratic is y`= ax^2+bx+c`

The function : `y = x^2-10x+2 " is in this form "`

with a = 1 , b = -10 and c = 2

the x-coord of vertex `= -b/(2a) = -(-10)/2 = 5`

now substitute x = 5 into equation to obtain y-coord

y-coord of vertex `= (5)^2 - 10(5) + 2 = 25-50+2 = -23`

thus vertex =( 5 , -23)

The axis of symmetry passes through the vertex and is parallel to the y-axis with equation x = 5

Here is the graph of the function with the axis of symmetry.
graph{(y-x^2+10x-2)(0.001y-x+5)=0 [-50.63, 50.6, -25.3, 25.32]}