Question

How do you find the vertex and axis of symmetry, and then graph the parabola given by: `y = x^2 - 5x + 3`?

Answer 1

`"vertex" = (5/2, -13/4); " axis of symmetry": x = 5/2`

Explanation:

Given: `y = x^2 -5x +3`

Standard form of a parabola: `y = a(x-k)^2 +k`

where #"vertex" = (h, k); "axis of symmetry": x = h"

and `a = " constant"`

The vertex can be found easily: `"vertex" = (-B/(2A), f(-B/(2A)))`

where the equation is in general form: `Ax^2 + Bx +C = 0`

`-B/(2A) = 5/2`

`f(-B/(2A)) = f(5/2) = (5/2)^2 -5(5/2) + 3`

`= 25/4-25/2 + 3 = 25/4 - 50/4 + 12/4 = -13/4`

`"vertex" = (5/2, -13/4)`

`" axis of symmetry": x = 5/2`

You can also do competing of the square to put the equation in standard form.

`y = (x^2 -5x) +3`

`y = (x - 5/2)^2 + 3 - (5/2)^2`

`y = (x - 5/2)^2 + 12/4 - 25/4`

`y = (x - 5/2)^2 - 13/4`

graph{x^2-5x+3 [-10, 10, -5, 5]}