# How do you find the axis of symmetry and vertex point of the function:  y=x^2-6x+4?

Oct 31, 2015

Axis of symmetry: $x = 3$
Vertex is at $\left(3 , - 5\right)$.

#### Explanation:

Notice that
x^2−6x+4 = x^2−6x+9 - 5 = (x-3)^2 - 5

Therefore, the graph of function
y = x^2−6x+4
can be obtained from the graph of function $y = {x}^{2}$ by shifting it by $3$ units to the right (that would be a graph of $y = {\left(x - 3\right)}^{2}$) and then shifting it by $5$ units down (to get to a graph of y = (x-3)^2−5).

graph{x^2-6x+4 [-7, 10, -7, 7]}

The above transformation shifts the axis of symmetry of function $y = {x}^{2}$ from the position coinciding with Y-axis to the right by $3$ units and it shifts the vertex of function $y = {x}^{2}$, firstly, to the right by $3$ and then down by $5$.

Now we know that the axis of symmetry is a vertical line parallel to Y-axis that intersects X-axis at point $x = 3$.

We also know that the vertex of the parabola is at point $\left(3 , - 5\right)$