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

May 14, 2018

$\text{vertex" = (5/2, -13/4); " axis of symmetry} : x = \frac{5}{2}$

#### Explanation:

Given: $y = {x}^{2} - 5 x + 3$

Standard form of a parabola: $y = a {\left(x - k\right)}^{2} + k$

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

and $a = \text{ constant}$

The vertex can be found easily: $\text{vertex} = \left(- \frac{B}{2 A} , f \left(- \frac{B}{2 A}\right)\right)$

where the equation is in general form: $A {x}^{2} + B x + C = 0$

$- \frac{B}{2 A} = \frac{5}{2}$

$f \left(- \frac{B}{2 A}\right) = f \left(\frac{5}{2}\right) = {\left(\frac{5}{2}\right)}^{2} - 5 \left(\frac{5}{2}\right) + 3$

$= \frac{25}{4} - \frac{25}{2} + 3 = \frac{25}{4} - \frac{50}{4} + \frac{12}{4} = - \frac{13}{4}$

$\text{vertex} = \left(\frac{5}{2} , - \frac{13}{4}\right)$

$\text{ axis of symmetry} : x = \frac{5}{2}$

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

$y = \left({x}^{2} - 5 x\right) + 3$

$y = {\left(x - \frac{5}{2}\right)}^{2} + 3 - {\left(\frac{5}{2}\right)}^{2}$

$y = {\left(x - \frac{5}{2}\right)}^{2} + \frac{12}{4} - \frac{25}{4}$

$y = {\left(x - \frac{5}{2}\right)}^{2} - \frac{13}{4}$

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