# How do you find the axis of symmetry and vertex point of the function: y=x^2-5x+3?

Oct 14, 2015

The axis of symmetry is $x = \frac{5}{2}$.
The vertex is $\left(\frac{5}{2} , - \frac{13}{4}\right)$.

#### Explanation:

$y = {x}^{2} - 5 x + 3$ is the standard form for a quadratic equation with the form $y = a x + b x + c$, where $a = 1 , b = - 5 , \mathmr{and} c = 3$. The graph of a quadratic equation is a parabola.

Axis of Symmetry
The axis of symmetry is the imaginary vertical line that divides the parabola into two equal halves.

For the standard form, the axis of symmetry is calculated as $x = \frac{- b}{2 a}$.

$x = \frac{- \left(- 5\right)}{2 \cdot 1} = \frac{5}{2}$

The axis of symmetry is $x = \frac{5}{2}$.

Vertex
The vertex is the minimum or maximum point of the parabola, and the $x$ value is the same as the axis of symmetry. Since $a$ is positive, the parabola opens upward, so the vertex is the minimum point.

To determine the $y$ value of the vertex, substitute the $x$ value of $\frac{5}{2}$ into the equation and solve for $y$.

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

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

Simplify.

$y = \frac{25}{4} - \frac{25}{2} + 3$

The LCD is $4$. Multiply the terms times numbers that will make the denominators $4$.

$y = \frac{25}{4} - \frac{25}{2} \times \frac{2}{2} + 3 \times \frac{4}{4} =$

$y = \frac{25}{4} - \frac{50}{4} + \frac{12}{4}$

Combine the numerators.

$y = \frac{25 - 50 + 12}{4} =$

$y = - \frac{13}{4}$

The vertex is $\left(\frac{5}{2} , - \frac{13}{4}\right)$.

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